Analysis I

Course C5

T. W. Korner

September 18, 2007

Small print The syllabus for the course is defined by the Faculty Board Schedules (which

are minimal for lecturing and maximal for examining). I should very much appreciate

being told of any corrections or possible improvements and might even part with a small

reward to the first finder of particular errors. This document is written in LATEX2e and

should be available from my home page http://www.dpmms.cam.ac.uk/˜twk in latex, dvi

and ps formats. My e-mail address is twk@dpmms. Some, fairly useless, comments on the

exercises are available for supervisors from me or the secretaries in DPMMS.

Contents

1 Why do we bother? 2

2 The axiom of Archimedes 3

3 Series and sums 4

4 Least upper bounds 7

5 Continuity 8

6 Differentiation 11

7 The mean value theorem 13

8 Complex variable 16

9 Power series 18

10 The standard functions 20

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11 Onwards to the complex plane 23

12 The Riemann integral 26

13 Some properties of the integral 29

14 Infinite integrals 32

15 Further reading 34

16 First example sheet 34

17 Second example sheet 39

18 Third example sheet 42

19 Fourth example sheet 53

1 Why do we bother?

It is surprising how many people think that analysis consists in the difficultproofs of obvious theorems. All we need know, they say, is what a limit is,the definition of continuity and the definition of the derivative. All the restis ‘intuitively clear’.

If pressed they will agree that these definitions apply as much to therationals Q as to the real numbers R. They then have to explain the followinginteresting example.

Example 1.1. If f : Q → Q is given by

f(x) = −1 if x2 < 2,

f(x) = 1 otherwise,

then(i) f is a continuous function with f(0) = −1, f(2) = 1 yet there does

not exist a c with f(c) = 0,(ii) f is a differentiable function with f ′(x) = 0 for all x yet f is not

constant.

What is the difference between R and Q which makes calculus work onone even though it fails on the other? Both are ‘ordered fields’, that is, bothsupport operations of ‘addition’ and ‘multiplication’ together with a relation‘greater than’ (‘order’) with the properties that we expect. If the reader is

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interested she will find a complete list of the appropriate axioms in texts likethe altogether excellent book of Spivak [5] and its many rather less excellentcompetitors, but, interesting as such things may be, they are irrelevant toour purpose which is not to consider the shared properties of R and Q but toidentify a difference between the two systems which will enable us to excludethe possibility of a function like that of Example 1.1 for functions from R toR.

To state the difference we need only recall a definition from course C3.

Definition 1.2. If an ∈ R for each n ≥ 1 and a ∈ R then we say that an → aif given ǫ > 0 we can find an n0(ǫ) such that

|an − a| < ǫ for all n ≥ n0(ǫ).

The key property of the reals, the fundamental axiom which makes ev-erything work was also stated in the course C3.

Axiom 1.3. [The fundamental axiom of analysis] If an ∈ R for eachn ≥ 1, A ∈ R and a1 ≤ a2 ≤ a3 ≤ . . . and an < A for each n then thereexists an a ∈ R such that an → a as n → ∞.

Less ponderously, and just as rigorously, the fundamental axiom for thereal numbers says every increasing sequence bounded above tends to a limit.

Everything which depends on the fundamental axiom is analysis, every-thing else is mere algebra.

2 The axiom of Archimedes

We start by proving the following results on limits, some of which you sawproved in course C3.

Lemma 2.1. (i) The limit is unique. That is, if an → a and an → b asn → ∞ then a = b.

(ii) If an → a as n → ∞ and n(1) < n(2) < n(3) . . . then an(j) → a asj → ∞.

(iii) If an = c for all n then an → c as n → ∞.(iv) If an → a and bn → b as n → ∞ then an + bn → a + b.(v) If an → a and bn → b as n → ∞ then anbn → ab.(vi) If an → a as n → ∞, an 6= 0 for each n and a 6= 0, then a−1

n → a−1.(vii) If an ≤ A for each n and an → a as n → ∞ then a ≤ A.

We need the following variation on the fundamental axiom.

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Exercise 2.2. A decreasing sequence of real numbers bounded below tends toa limit.

[Hint. If a ≤ b then −b ≤ −a.]Useful as the results of Lemma 2.1 are, they are also true of sequences in

Q. They are therefore mere, if important, algebra. Our first truly ‘analysis’result may strike the reader as rather odd.

Theorem 2.3. [Axiom of Archimedes]

1

n→ 0 as n → ∞

Theorem 2.3 shows that there is no ‘exotic’ real number ג say (to choosean exotic symbol) with the property that 1/n > ג for all integers n ≥ 1 yetג > 0 (that is, ג is strictly positive and yet smaller than all strictly positiverationals). There exist number systems with such exotic numbers (the famous‘non-standard analysis’ of Abraham Robinson and the ‘surreal numbers’ ofConway constitute two such systems) but, just as the rationals are, in somesense, too small a system for the standard theorems of analysis to hold sothese non-Archimedean systems are, in some sense, too big. Archimedesand Eudoxus realised the need for an axiom to show that there is no exoticnumber k bigger than any integer1 (i.e. k > n for all integers n ≥ 1; to seethe connection with our form of the axiom consider ג = 1/k). However, inspite of its name, what was an axiom for Archimedes is a theorem for us.

Theorem 2.4. Given any real number K we can find an integer n withn > K.

3 Series and sums

There is no need to restrict the notion of a limit to real numbers.

Definition 3.1. If an ∈ C for each n ≥ 1 and a ∈ C then we say that an → aif given ǫ > 0 we can find an n0(ǫ) such that

|an − a| < ǫ for all n ≥ n0(ǫ).

Exercise 3.2. We work in C.(i) The limit is unique. That is, if an → a and an → b as n → ∞ then

a = b.

1Footnote for passing historians, this is a course in mathematics.

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(ii) If an → a as n → ∞ and n(1) < n(2) < n(3) . . . then an(j) → a asj → ∞.

(iii) If an = c for all n then an → c as n → ∞.(iv) If an → a and bn → b as n → ∞ then an + bn → a + b.(v) If an → a and bn → b as n → ∞ then anbn → ab.(vi) If an → a as n → ∞, an 6= 0 for each n and a 6= 0, then a−1

n → a−1.

Exercise 3.3. Explain why there is no result in Exercise 3.2 correspondingto part (vii) of Lemma 2.1.

We illustrate some of the ideas introduced by studying infinite sums.

Definition 3.4. We work in F where F = R or F = C. If aj ∈ F we saythat

∑∞

j=1 aj converges to s if

N∑

j=1

aj → s

as N → ∞. We write∑∞

j=1 aj = s.

If∑N

j=1 aj does not tend to a limit as N → ∞, we say that the sum∑∞

j=1 aj diverges.

Lemma 3.5. We work in F where F = R or F = C.(i) Suppose aj, bj ∈ F and λ, µ ∈ F. If

∑∞

j=1 aj and∑∞

j=1 bj convergethen so does

∑∞

j=1 λaj + µbj and

∞∑

j=1

λaj + µbj = λ∞∑

j=1

aj + µ∞∑

j=1

bj.

(ii) Suppose aj, bj ∈ F and there exists an N such that aj = bj for allj ≥ N . Then, either

∑∞

j=1 aj and∑∞

j=1 bj both converge or they both diverge.(In other words, initial terms do not matter.)

Exercise 3.6. Any problem on sums∑∞

j=1 aj can be converted into one onsequences by considering the sequence sn =

∑nj=1 aj. Show conversely that a

sequence sn converges if and only if, when we set a1 = s1 and an = sn − sn−1

[n ≥ 2] we have∑∞

j=1 aj convergent. What can you say about limn→∞

∑nj=1 aj

and limn→∞ sn if both exist?

The following results are fundamental to the study of sums.

Theorem 3.7. (The comparison test.) We work in R. Suppose that0 ≤ bj ≤ aj for all j. Then, if

∑∞

j=1 aj converges, so does∑∞

j=1 bj.

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Theorem 3.8. We work in F where F = R or F = C. If∑∞

j=1 |aj| converges,then so does

∑∞

j=1 aj.

Theorem 3.8 is often stated using the following definition.

Definition 3.9. We work in F where F = R or F = C. If∑∞

j=1 |aj| convergeswe say that the sum

∑∞

j=1 aj is absolutely convergent.

Theorem 3.8 then becomes the statement that absolute convergence im-plies convergence.

Here is a trivial but useful consequence of Theorems 3.7 and 3.8.

Lemma 3.10. (Ratio test.) Suppose aj ∈ C and |aj+1/aj| → l as j → ∞.If l < 1, then

∑∞

j=1 |aj| converges. If l > 1, then∑∞

j=1 aj diverges.

Of course Lemma 3.10 tells us nothing if l = 1 or l does not exist.Sums which are not absolutely convergent are much harder to deal with

in general. It is worth keeping in mind the following trivial observation.

Lemma 3.11. We work in F where F = R or F = C. If∑∞

j=1 aj converges,then aj → 0 as j → ∞.

At a deeper level the following result is sometimes useful.

Lemma 3.12. (Alternating series test.) We work in R. If we have adecreasing sequence of positive numbers an with an → 0 as n → ∞, then∑∞

j=1(−1)j+1aj converges.Further ∣

∣∣∣∣

N∑

j=1

(−1)j+1aj −∞∑

j=1

(−1)j+1aj

∣∣∣∣∣≤ |aN+1|

for all N ≥ 1.

The last sentence is sometimes expressed by saying ‘the error caused byreplacing a convergent infinite alternating sum of decreasing terms by thesum of its first few terms is no greater than the absolute value of the firstterm neglected’.

Later we will give another test for convergence called the integral test(Lemma 14.4) from which we deduce the result known to many of you that∑∞

n=1 n−1 diverges. I will give another proof in the next example.

Example 3.13. (i)∞∑

n=1

1

ndiverges.

(ii)∞∑

n=1

(−1)n

nis convergent but not absolutely convergent.

(iii) If v2n = 1/n, v2n−1 = −1/(2n) then∑∞

n=1 vn is not convergent.

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4 Least upper bounds

A non-empty bounded set in R need not have a maximum.

Example 4.1. The set E = {−1/n : n ≥ 1} is non-empty and any e ∈ Esatisfies the inequalities −1 ≤ e ≤ 0 but E has no largest member.

However, as we shall see every non-empty bounded set in R has a leastupper bound (or supremum).

Definition 4.2. Let E be a non-empty set in R. We say that α is a leastupper bound for E if

(i) α ≥ e for all e ∈ E [that is, α is an upper bound for E] and(ii) If β ≥ e for all e ∈ E then β ≥ α [that is, α is the least such upper

bound].

If E has a supremum α we write supe∈E e = sup E = α.

Lemma 4.3. If the least upper bound exists it is unique.

The following remark is trivial but sometimes helpful.

Lemma 4.4. Let E be a non-empty set in R. Then α is a least upper boundfor E if and only if we can find en ∈ E with en → α and bn such that bn ≥ efor all e ∈ E and bn → α as n → ∞.

Here is the promised result.

Theorem 4.5. Any non-empty set in R with an upper bound has a leastupper bound.

We observe that this result is actually equivalent to the fundamentalaxiom.

Theorem 4.6. Theorem 4.5 implies the fundamental axiom.

Of course we have the notion of a greatest lower bound or infimum.

Exercise 4.7. Define the greatest lower bound in the manner of Defini-tion 4.2, prove its uniqueness in the manner of Lemma 4.3 and state andprove a result corresponding to Lemma 4.4.

If E has an infimum β we write infe∈E e = inf E = β. One way of dealingwith the infimum is to use the following observation.

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Lemma 4.8. Let E be a non-empty set in R and write −E = {−e : e ∈ E}.Then E has an infimum if and only if −E has a supremum. If E has aninfimum inf E = − sup(−E).

Exercise 4.9. Use Lemma 4.8 and Theorem 4.5 to show that any non-emptyset in R with a lower bound has a greatest lower bound.

The notion of a supremum will play an important role in our proofs ofTheorem 5.12 and Theorem 9.2.

The following result is also equivalent to the fundamental axiom (that is,we can deduce it from the fundamental axiom and conversely, if we take itas an axiom, rather than a theorem, then we can deduce the fundamentalaxiom as a theorem).

Theorem 4.10. [Bolzano-Weierstrass] If xn ∈ R and there exists a Ksuch that |xn| ≤ K for all n, then we can find n(1) < n(2) < . . . and x ∈ R

such that xn(j) → x as j → ∞.

The Bolzano-Weierstrass theorem says that every bounded sequence of re-als has a convergent subsequence. Notice that we say nothing about unique-ness; if xn = (−1)n then x2n → 1 but x2n+1 → −1 as n → ∞.

We shall prove the theorem of Bolzano-Weierstrass by ‘lion hunting’but your supervisor may well show you another method. We shall use theBolzano-Weierstrass theorem to prove that every continuous function on aclosed bounded interval is bounded and attains its bounds (Theorem 5.12).The Bolzano-Weierstrass theorem will be much used in the next analysiscourse because it generalises to many dimensions.

5 Continuity

We make the following definition.

Definition 5.1. A function f : R → R is continuous at x if given ǫ > 0 wecan find a δ(ǫ, x) > 0 [read ‘a delta depending on epsilon and x’] such that

|f(x) − f(y)| < ǫ

for all y with |x − y| < δ(ǫ, x).If f is continuous at each point x ∈ R we say that f is a continuous

function on R.

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I shall do my best to make this seem a reasonable definition but it isimportant to realise that I am really stating a rule of the game (like a knightsmove in chess or the definition of offside in football). If you wish to play thegame you must accept the rules. Results about continuous functions mustbe derived from the definition and not stated as ‘obvious from the notion ofcontinuity’.

In practice we use a slightly more general definition.

Definition 5.2. Let E be a subset of R. A function f : E → R is continuousat x ∈ E if given ǫ > 0 we can find a δ(ǫ, x) > 0 [read ‘a delta depending onepsilon and x’] such that

|f(x) − f(y)| < ǫ

for all y ∈ E with |x − y| < δ(ǫ, x).If f is continuous at each point x ∈ E, we say that f is a continuous

function on E.

However, it will do no harm and may be positively helpful if, whilst youare getting used to the idea of continuity, you concentrate on the case E = R.

Lemma 5.3. Suppose that E is a subset of R, that x ∈ E, and that f and gare functions from E to R.

(i) If f(x) = c for all x ∈ E, then f is continuous on E.(ii) If f and g are continuous at x, then so is f + g.(iii) Let us define f × g : E → R by f × g(t) = f(t)g(t) for all t ∈ E.

Then if f and g are continuous at x, so is f × g.(iv) Suppose that f(t) 6= 0 for all t ∈ E. If f is continuous at x so is

1/f .

Lemma 5.4. Let U and V be subsets of R. Suppose f : U → R is suchthat f(t) ∈ V for all t ∈ U . If f is continuous at x ∈ U and g : V → R iscontinuous at f(x), then the composition g ◦ f is continuous at x.

By repeated use of parts (ii) and (iii) of Lemma 5.3 it is easy to showthat polynomials P (t) =

∑nr=0 art

r are continuous. The details are spelledout in the next exercise.

Exercise 5.5. Prove the following results.(i) Suppose that E is a subset of R and that f : E → R is continuous at

x ∈ E. If x ∈ E ′ ⊂ E then the restriction f |E′ of f to E ′ is also continuousat x.

(ii) If J : R → R is defined by J(x) = x for all x ∈ R, then J iscontinuous on R.

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(iii) Every polynomial P is continuous on R.(iv) Suppose that P and Q are polynomials and that Q is never zero on

some subset E of R. Then the rational function P/Q is continuous on E(or, more precisely, the restriction of P/Q to E is continuous.)

The following result is little more than an observation but will be veryuseful.

Lemma 5.6. Suppose that E is a subset of R, that x ∈ E, and that fis continuous at x. If xn ∈ E for all n and xn → x as n → ∞, thenf(xn) → f(x) as n → ∞.

So far in this section we have only done algebra but the next result de-pends on the fundamental axiom. It is one of the key results of analysisand although my recommendation runs contrary to a century of enlightenedpedagogy I can see no objections to students learning the proof as a model.Notice that the theorem resolves the problem posed by Example 1.1 (i).

Theorem 5.7. (The intermediate value theorem). If f : [a, b] → R

is continuous and f(a) ≥ 0 ≥ f(b) then there exists a c ∈ [a, b] such thatf(c) = 0.

Exercises 5.8 to 5.10 are applications of the intermediate value theorem.

Exercise 5.8. Show that any real polynomial of odd degree has at least oneroot. Is the result true for polynomials of even degree? Give a proof orcounterexample.

Exercise 5.9. Suppose that g : [0, 1] → [0, 1] is a continuous function. Byconsidering f(x) = g(x) − x, or otherwise, show that there exists a c ∈ [0, 1]with g(c) = c. (Thus every continuous map of [0, 1] into itself has a fixedpoint.) Give an example of a bijective (but, necessarily, non-continuous)function h : [0, 1] → [0, 1] such that h(x) 6= x for all x ∈ [0, 1].[Hint: First find a function H : [0, 1] \ {0, 1, 1/2} → [0, 1] \ {0, 1, 1/2}such that H(x) 6= x.]

Exercise 5.10. Every mid-summer day at six o’clock in the morning, theyoungest monk from the monastery of Damt starts to climb the narrow pathup Mount Dipmes. At six in the evening he reaches the small temple at thepeak where he spends the night in meditation. At six o’clock in the morningon the following day he starts downwards, arriving back at the monastery atsix in the evening. Of course, he does not always walk at the same speed.Show that, none the less, there will be some time of day when he will be atthe same place on the path on both his upward and downward journeys.

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We proved the intermediate value theorem (Theorem 5.7) by lion hunting.We prove the next two theorems by using the Bolzano-Weierstrass Theorem(Theorem 4.10). Again the results are very important and I can see noobjection to learning the proofs as a model.

Theorem 5.11. If f : [a, b] → R is continuous then we can find an M suchthat |f(x)| ≤ M for all x ∈ [a, b].

In other words a continuous function on a closed bounded interval isbounded. We improve this result in the next theorem.

Theorem 5.12. If f : [a, b] → R is continuous then we can find x1, x2 ∈ [a, b]such that

f(x1) ≤ f(x) ≤ f(x2)

for all x ∈ [a, b].

In other words a continuous function on a closed bounded interval isbounded and attains its bounds.

6 Differentiation

In this section it will be useful to have another type of limit.

Definition 6.1. Let E be a subset of R, f be some function from E to R,and x some point of E. If l ∈ R we say that f(y) → l as y → x [or, if wewish to emphasise the restriction to E that f(y) → l as y → x through valuesy ∈ E] if, given ǫ > 0, we can find a δ(ǫ) > 0 [read ‘a delta depending onepsilon’] such that

|f(y) − l| < ǫ

for all y ∈ E with 0 < |x − y| < δ(ǫ).

As before there is no real loss if the reader initially takes E = R.The following two exercises are easy but useful.

Exercise 6.2. Let E be a subset of R. Show that a function f : E → R iscontinuous at x ∈ E if and only if f(y) → f(x) as y → x.

Exercise 6.3. Let E be a subset of R, f , g be some functions from E to R,and x some point of E. (i) The limit is unique. That is, if f(y) → l andf(y) → k as y → x then l = k.

(ii) If x ∈ E ′ ⊆ E and f(y) → l as y → x through values y ∈ E, thenf(y) → l as y → x through values y ∈ E ′.

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(iii) If f(t) = c for all t ∈ E then f(y) → c as y → x.(iv) If f(y) → l and g(y) → k as y → x then f(y) + g(y) → l + k.(v) If f(y) → l and g(y) → k as y → x then f(y)g(y) → lk.(vi) If f(y) → l as y → x, f(t) 6= 0 for each t ∈ E and l 6= 0 then

f(t)−1 → l−1.(vii) If f(t) ≤ L for each t ∈ E and f(y) → l as y → x then l ≤ L.

We can now define the derivative.

Definition 6.4. Let E be a subset of R. A function f : E → R is differen-tiable at x ∈ E with derivative f ′(x) if

f(y) − f(x)

y − x→ f ′(x)

as y → x.If f is differentiable at each point x ∈ E, we say that f is a differentiable

function on E.

As usual, no harm will be done if you replace E by R.Here are some easy consequences of the definition.

Exercise 6.5. Let E be a subset of R, f some function from E to R, and xsome point of E. Show that if f is differentiable at x then f is continuousat x.

Exercise 6.6. Let E be a subset of R, f , g be some functions from E to R,and x some point of E. Prove the following results.

(i) If f(t) = c for all t ∈ E then f is differentiable at x with f ′(x) = 0.(ii) If f and g are differentiable at x then so is their sum f + g and

(f + g)′(x) = f ′(x) + g′(x).

(iii) If f and g are differentiable at x then so is their product f × g and

(f × g)′(x) = f ′(x)g(x) + f(x)g′(x).

(iv) If f is differentiable at x and f(t) 6= 0 for all t ∈ E then 1/f isdifferentiable at x and

(1/f)′(x) = −f ′(x)/f(x)2.

(v) If f(t) =∑N

r=0 artr on E then f is differentiable at x and

f ′(x) =N∑

r=1

rarxr−1

.

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The next result is slightly harder to prove than it looks. (we split theproof into two halves depending on whether f ′(x) 6= 0 or f ′(x) = 0.

Lemma 6.7. [Chain rule] Let U and V be subsets of R. Suppose f : U → R

is such that f(t) ∈ V for all t ∈ U . If f is differentiable at x ∈ U andg : V → R is differentiable at f(x), then the composition g◦f is differentiableat x with

(g ◦ f)′(x) = f ′(x)g′(f(x)).

7 The mean value theorem

We have almost finished our project of showing that the horrid situationrevealed by Example 1.1 can not occur for the reals.

Our first step is to prove Rolle’s theorem.

Theorem 7.1. [Rolle’s theorem] If g : [a, b] → R is a continuous functionwith g differentiable on (a, b) and g(a) = g(b), then we can find a c ∈ (a, b)such that g′(c) = 0.

A simple tilt gives the famous mean value theorem.

Theorem 7.2. (The mean value theorem). If f : [a, b] → R is a con-tinuous function with f differentiable on (a, b), then we can find a c ∈ (a, b)such that

f(b) − f(a) = (b − a)f ′(c).

We now have the results so long desired.

Lemma 7.3. If f : [a, b] → R is a continuous function with f differentiableon (a, b), then the following results hold.

(i) If f ′(t) > 0 for all t ∈ (a, b) then f is strictly increasing on [a, b].(That is, f(y) > f(x) whenever b ≥ y > x ≥ a.)

(ii) If f ′(t) ≥ 0 for all t ∈ (a, b) then f is increasing on [a, b]. (That is,f(y) ≥ f(x) whenever b ≥ y > x ≥ a.)

(iii) [The constant value theorem] If f ′(t) = 0 for all t ∈ (a, b) thenf is constant on [a, b]. (That is, f(y) = f(x) whenever b ≥ y > x ≥ a.)

Notice that since we deduce Lemma 7.3 from the mean value theorem wecan not use it in the proof of Rolle’s theorem.

The mean value theorem has many important consequences, some ofwhich we look at in the remainder of the section.

We start by looking at inverse functions.

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Lemma 7.4. (i) Suppose f : [a, b] → R is continuous. Then f is injectiveif and only if it is strictly increasing (that is f(t) > f(s) whenever a ≤ s <t ≤ b) or strictly decreasing.

(ii) Suppose f : [a, b] → R is continuous and strictly increasing. Letf(a) = c and f(b) = d. Then the map f : [a, b] → [c, d] is bijective and f−1

is continuous on [c, d].

Lemma 7.5. [Inverse rule] Suppose f : [a, b] → R is differentiable on [a, b]and f ′(x) > 0 for all x ∈ [a, b]. Let f(a) = c and f(b) = d. Then the mapf : [a, b] → [c, d] is bijective and f−1 is differentiable on [c, d] with

f−1′(x) =1

f ′(f−1(x)).

In the opinion of the author the true meaning of the inverse rule andthe chain rule only becomes clear when we consider higher dimensions in thenext analysis course.

We now prove a form of Taylor’s theorem.

Theorem 7.6. [nth mean value theorem] Suppose that b > a and f :[a, b] → R is n + 1 times differentiable. Then

f(b) −n∑

j=0

f (j)(a)

j!(b − a)j =

f (n+1)(c′)

(n + 1)!(b − a)n+1

for some c′ with a < c′ < b.

This gives us a global and a local Taylor’s theorem.

Theorem 7.7. [Global Taylor theorem] Suppose that b > a and f :[a, b] → R is n + 1 times differentiable. If x, t ∈ [a, b]

f(t) =n∑

j=0

f (j)(x)

j!(t − x)j +

f (n+1)(x + θ(t − x))

(n + 1)!(t − x)n+1

for some θ ∈ (0, 1).

Theorem 7.8. [Local Taylor theorem] Suppose that δ > 0 and f : (x −δ, x+ δ) → R is n times differentiable on (x− δ, x+ δ) and f (n) is continuousat x. Then

f(t) =n∑

j=0

f (j)(x)

j!(t − x)j + ǫ(t)(t − x)n

where ǫ(t) → 0 as t → x.

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Notice that the local Taylor theorem always gives us some informationbut that the global one is useless unless we can find a useful bound on then + 1th derivative.

To reinforce this warning we consider a famous example of Cauchy.

Exercise 7.9. Consider the function F : R → R defined by

F (0) = 0

F (x) = exp(−1/x2) otherwise.

(i) Prove by induction, using the standard rules of differentiation, that Fis infinitely differentiable at all points x 6= 0 and that, at these points,

F (n)(x) = Pn(1/x) exp(−1/x2)

where Pn is a polynomial which need not be found explicitly.(ii) Explain why x−1Pn(1/x) exp(−1/x2) → 0 as x → 0.(iii) Show by induction, using the definition of differentiation, that F is

infinitely differentiable at 0 with F (n)(0) = 0 for all n. [Be careful to get thispart of the argument right.]

(iv) Show that

F (x) =∞∑

j=0

F (j)(0)

j!xj

if and only if x = 0. (The reader may prefer to say that ‘The Taylor expansionof F is only valid at 0’.)

(v) Why does part (iv) not contradict the local Taylor theorem (Theo-rem 7.8)?

Since examiners are fonder of the global Taylor theorem than it deservesI shall go through the following example.

Example 7.10. Assuming the standard properties of the exponential functionshow that

exp x =∞∑

j=0

xj

j!

for all x.

Please note that in a pure mathematics question many (or even most) ofthe marks in a question of this type will depend on estimating the remainderterm. [In methods questions you may simply be asked to ‘find the Taylor’sseries’ without being asked to prove convergence.]

15

8 Complex variable

The field C of complex numbers resembles the field R of real numbers inmany ways but not in all.

Lemma 8.1. We can not define an order on C which will behave in the sameway as > for R.

However there is sufficient similarity for us to define limits, continuityand differentiability. (We have already seen some of this in Definition 3.1and Exercise 3.2.)

Definition 8.2. Let E be a subset of C, f be some function from E to C,and z some point of E. If l ∈ C we say that f(w) → l as w → z [or, ifwe wish to emphasise the restriction to E that f(w) → l as w → z throughvalues w ∈ E] if, given ǫ > 0, we can find a δ(ǫ) > 0 [read ‘a delta dependingon epsilon’] such that

|f(w) − l| < ǫ

for all w ∈ E with 0 < |w − z| < δ(ǫ).

As usual there is no real loss if the reader initially takes E = C.

Definition 8.3. Let E be a subset of C. We say that a function f : E → C

is continuous at z ∈ E if and only if f(w) → f(z) as w → z.

Exercise 8.4. Let E be a subset of C, f , g be some functions from E to C,and z some point of E.

(i) The limit is unique. That is, if f(w) → l and f(w) → k as w → zthen l = k.

(ii) If z ∈ E ′ ⊆ E and f(w) → l as w → z through values w ∈ E, thenf(w) → l as w → x through values w ∈ E ′.

(iii) If f(u) = c for all u ∈ E then f(w) → c as w → z.(iv) If f(w) → l and g(w) → k as w → z then f(w) + g(w) → l + k.(v) If f(w) → l and g(w) → k as w → z then f(w)g(w) → lk.(vi) If f(w) → l as w → z, f(u) 6= 0 for each u ∈ E and l 6= 0 then

f(w)−1 → l−1.

Exercise 8.5. Suppose that E is a subset of C, that z ∈ E, and that f andg are functions from E to C.

(i) If f(u) = c for all u ∈ E, then f is continuous on E.(ii) If f and g are continuous at z, then so is f + g.(iii) Let us define f × g : E → C by f × g(u) = f(u)g(u) for all u ∈ E.

Then if f and g are continuous at z, so is f × g.

16

(iv) Suppose that f(u) 6= 0 for all u ∈ E. If f is continuous at z so is1/f .

(v) If z ∈ E ′ ⊂ E and f is continuous at z then the restriction f |E′ of fto E ′ is also continuous at z.

(vi) If J : C → C is defined by J(z) = z for all z ∈ C, then J iscontinuous on C.

(vii) Every polynomial P is continuous on C.(viii) Suppose that P and Q are polynomials and that Q is never zero on

some subset E of C. Then the rational function P/Q is continuous on E(or, more precisely, the restriction of P/Q to E is continuous.)

Exercise 8.6. Let U and V be subsets of C. Suppose f : U → C is suchthat f(z) ∈ V for all z ∈ U . If f is continuous at w ∈ U and g : V → C iscontinuous at f(w), then the composition g ◦ f is continuous at w.

Definition 8.7. Let E be a subset of C. A function f : E → C is differen-tiable at z ∈ E with derivative f ′(z) if

f(w) − f(z)

w − z→ f ′(z)

as w → z.If f is differentiable at each point z ∈ E we say that f is a differentiable

function on E.

Exercise 8.8. Let E be a subset of C, f some function from E to C, and zsome point of E. Show that if f is differentiable at z then f is continuous atz.

Exercise 8.9. Let E be a subset of C, f , g be some functions from E to C,and z some point of E. Prove the following results.

(i) If f(u) = c for all u ∈ E then f is differentiable at z with f ′(z) = 0.(ii) If f and g are differentiable at z then so is their sum f + g and

(f + g)′(z) = f ′(z) + g′(z).

(iii) If f and g are differentiable at z then so is their product f × g and

(f × g)′(z) = f ′(z)g(z) + f(z)g′(z).

(iv) If f is differentiable at z and f(u) 6= 0 for all u ∈ E then 1/f isdifferentiable at z and

(1/f)′(z) = −f ′(z)/f(z)2.

17

(v) If f(u) =∑N

r=0 arur on E then f is differentiable at z and

f ′(z) =N∑

r=1

rarzr−1

.

Exercise 8.10. [Chain rule] Let U and V be subsets of C. Suppose f : U →C is such that f(z) ∈ V for all t ∈ U . If f is differentiable at w ∈ U andg : V → R is differentiable at f(w), then the composition g◦f is differentiableat w with

(g ◦ f)′(w) = f ′(w)g′(f(w)).

In spite of these similarities the subject of complex differentiable functionsis very different from that of real differentiable functions. It turns out that‘well behaved’ complex functions need not be differentiable.

Example 8.11. Consider the map Γ : C → C given by Γ(z) = z∗. Thefunction Γ is nowhere differentiable.

Because complex differentiability is so much more restrictive than realdifferentiability we can prove stronger theorems about complex differentiablefunctions. For example it can be shown that such functions can be writtenlocally as power series2 (contrast the situation in the real case revealed byExample 7.9). To learn more go to the course P3 on complex methods.

9 Power series

In this section we work in C unless otherwise stated. We start with a veryuseful observation.

Lemma 9.1. If∑∞

n=0 anzn0 converges and |z| < |z0| then

∑∞

n=0 anzn con-

verges.

This gives us the following basic theorem on power series.

Theorem 9.2. Suppose that an ∈ C. Then either∑∞

n=0 anzn converges for

all z ∈ C, or there exists a real number R with R ≥ 0 such that(i)∑∞

n=0 anzn converges if |z| < R,

(ii)∑∞

n=0 anzn diverges if |z| > R.

2The syllabus says that this fact is part of this course. In this one instance I adviseyou to ignore the syllabus.

18

We call R the radius of convergence of∑∞

n=0 anzn. If

∑∞

n=0 anzn converges

for all z we write R = ∞. The following useful strengthening is left to thereader as an exercise.

Exercise 9.3. Suppose that∑∞

n=0 anzn has radius of convergence R. Then

the sequence |anzn| is unbounded if |z| > R and

∑∞

n=0 anzn converges abso-lutely if |z| < R.

Note that we say nothing about what happens on the circle of conver-gence.

Example 9.4. (i)∑∞

n=1 n−2zn has radius of convergence 1 and convergesfor all z with |z| = 1.

(ii)∑∞

n=1 zn has radius of convergence 1 and diverges for all z with |z| =1.

A more complicated example is given in Exercise 18.16.It is a remarkable fact that we can operate with power series in the same

way as polynomials (within the radius of convergence). In particular we shallshow that we can differentiate term by term.

Theorem 9.5. If∑∞

n=0 anzn has radius of convergence R and we write

f(z) =∑∞

n=0 anzn then f is differentiable at all points z with |z| < R and

f ′(z) =∞∑

n=1

nanzn−1.

The proof is starred in the syllabus. We use three simple observations.

Lemma 9.6. If∑∞

n=0 anzn has radius of convergence R then given any ǫ > 0

we can find a K(ǫ) such that∑∞

n=0 |anzn| < K(ǫ) for all |z| ≤ R − ǫ.

Lemma 9.7. If∑∞

n=0 anzn has radius of convergence R then so do

∑∞

n=1 nanzn−1

and∑∞

n=2 n(n − 1)anzn−2.

Lemma 9.8. (i)

(n

r

)

≤ n(n − 1)

(n − 2

r − 2

)

for all 2 ≤ r ≤ n.

(ii) |(z + h)n − zn −nhzn−1| ≤ n(n− 1)(|z|+ |h|)n−2|h|2 for all z, h ∈ C.

In this course we shall mainly work on the real line. Restricting to thereal line we obtain the following result.

19

Theorem 9.9. (i) If aj ∈ R there exists a unique R (the radius of conver-gence) with 0 ≤ R ≤ ∞ such that

∑∞

n=0 anxn converges for all real x with

|x| < R and diverges for all real x with x > R.(ii) If

∑∞

n=1 anxn has radius of convergence R and we write f(x) =

∑∞

n=0 anxn then f is differentiable at all points x with |x| < R and

f ′(x) =∞∑

n=1

nanxn−1.

10 The standard functions

In school you learned all about the functions exp, log, sin and cos and aboutthe behaviour of xα. Nothing that you learned was wrong (we hope) but youmight be hard pressed to prove all the facts you know in a coherent manner,

To get round this problem, we start from scratch making new definitionsand assuming nothing about these various functions. One of your tasks isto make sure that the lecturer does not slip in some unproved fact. On theother hand you must allow your lecturer to choose definitions which allow aneasy development of the subject rather than those that follow some ‘historic’,‘intuitive’ or ‘pedagogically appropriate’ path3.

Let us start with the exponential function. Throughout this section weshall restrict ourselves to the real line.

Lemma 10.1. The sum∑∞

n=0xn

n!has infinite radius of convergence.

We can thus define a function e : R → R by

e(x) =∞∑

n=0

xn

n!.

(We use e(x) rather than exp(x) to help us avoid making unjustified assump-tions.)

Theorem 10.2. (i) The function e : R → R is everywhere differentiable withe′(x) = e(x).

(ii) e(x + y) = e(x)e(y) for all x, y ∈ R.(iii) e(x) > 0 for all x ∈ R.(iv) e is a strictly increasing function.(v) e(x) → ∞ as x → ∞, e(x) → 0 as x → −∞.(vi) e : R → (0,∞) is a bijection.

3If you want to see a treatment along these lines see the excellent text of Burn[3].

20

It is worth stating some of our results in the language of group theory.

Lemma 10.3. The mapping e is an isomorphism of the group (R, +) andthe group ((0,∞),×).

Since e : R → (0,∞) is a bijection we can consider the inverse functionl : (0,∞) → R.

Theorem 10.4. (i) l : (0,∞) → R is a bijection. We have l(e(x)) = x forall x ∈ R and e(l(t)) = t for all t ∈ (0,∞).

(ii) The function l : (0,∞) → R is everywhere differentiable with l′(t) =1/t.

(iii) l(uv) = l(u) + l(v) for all u, v ∈ (0,∞).

We now write e(x) = exp(x), l(t) = log t (the use of ln is unnecessary).If α ∈ R and x > 0. we write rα(x) = exp(α log x).

Theorem 10.5. Suppose x, y > 0 and α, β ∈ R. Then(i) rα(xy) = rα(x)rα(y),(ii) rα+β(x) = rα(x)rβ(x),(iii) rα(rβ(x)) = rαβ(x)(iv) r1(x) = x.

Exercise 10.6. Use the results of Theorem 10.5 to show that if n is a strictlypositive integer and x > 0 then

rn(x) = xx . . . x︸ ︷︷ ︸

n

.

Thus, if we write xα = rα(x), our new notation is consistent with our oldschool notation when α is rational but gives, in addition, a valid definitionwhen α is irrational.

Lemma 10.7. (i) If α is real, rα : (0,∞) → (0,∞) is everywhere differen-tiable and r′α(x) = αrα−1(x).

(ii) If x > 0 and we define fx(α) = xα then fx : R → (0,∞) is everywheredifferentiable and f ′

x(α) = log xfx(α).

Finally we look at the trigonometric functions.

Lemma 10.8. The sums∑∞

n=0(−1)nx2n+1

(2n+1)!and

∑∞

n=0(−1)nx2n

(2n)!have infinite

radius of convergence.

We can thus define functions s, c : R → R by

s(x) =∞∑

n=0

(−1)nx2n+1

(2n + 1)!and c(x) =

∞∑

n=0

(−1)nx2n

(2n)!

21

Lemma 10.9. (i) The functions s, c : R → R are everywhere differentiablewith s′(x) = c(x) and c′(x) = −s(x).

(ii) s(x + y) = s(x)c(y) + c(x)s(y), c(x + y) = c(x)c(y) − s(x)s(y) ands(x)2 + c(x)2 = 1 for all x, y ∈ R.

(iii) s(−x) = −s(x), c(−x) = c(x) for all x ∈ R.

Exercise 10.10. Write down what you consider to be the chief properties ofsinh and cosh. (They should convey enough information to draw a reasonablegraph of the two functions.)

(i) Obtain those properties by starting with the definitions

sinh x =∞∑

n=0

x2n+1

(2n + 1)!and cosh x =

∞∑

n=0

x2n

(2n)!

and proceeding along the lines of Lemma 10.9.(ii) Obtain those properties by starting with the definitions

sinh x =exp x − exp(−x)

2and cosh x =

exp x + exp(−x)

2

We have not yet proved one of the most remarkable properties of the sineand cosine functions, their periodicity.

Theorem 10.11. Let s and c be as in Lemma 10.9.(i) If c(a) = 0 and c(b) = 0 then s(b − a) = 0.(ii) We have s(x) > 0 for all 0 < x ≤ 1.(iii) There exists a unique ω ∈ [0, 2] such that c(ω) = 0.(iv) s(ω) = 1.(v) s(x + 4ω) = s(x), c(x + 4ω) = c(ω).(vi) The function c is strictly decreasing from 1 to −1 as x runs from 0

to 2ω,

We now define π = 2ω. If x and y are non-zero vectors in Rm we knowby the Cauchy-Schwarz inequality that |x.y| ≤ ‖x‖‖y‖ and we may definethe angle between the two vectors to be that θ with 0 ≤ θ ≤ π which satisfies

cos θ =x.y

‖x‖‖y‖ .

We can also justify the standard use of polar coordinates.

Lemma 10.12. If (x, y) ∈ R2 and (x, y) 6= (0, 0) then there exist a uniquer > 0 and θ with 0 ≤ θ < 2π such that

x = r cos θ, y = r sin θ.

22

You may be unhappy with a procedure which reduces geometry to analy-sis. It is possible to produce treatments which soften the blow4 but what wecan not do is to justify analytic results by appealing to geometry and thenappeal to analysis to justify the geometry.

11 Onwards to the complex plane

This section contains useful background material which does not really formpart of the course. If time is short I shall omit it entirely.

The mean value theorem fails for differentiable functions f : C → C.(See Example 11.5.) However, the constant value theorem holds.

Theorem 11.1. If f : C → C is differentiable and f ′(z) = 0 for all z ∈ C

then f is constant.

The proof of Theorem 11.1 that I shall give is very ad hoc and you willmeet better ones later.

Since the constant value theorem holds we can extend the proof of The-orem 10.2 (ii) to this wider context and obtain a version of the exponentialfunction for complex numbers. In this section we work in C unless otherwisestated.

Lemma 11.2. The sum∑∞

n=0zn

n!has infinite radius of convergence.

We can thus define a function exp : C → C by

exp(z) =∞∑

n=0

zn

n!.

Theorem 11.3. (i) The function e : C → C is everywhere differentiable withe′(z) = e(z).

(ii) e(z + w) = e(z)e(w) for all z, w ∈ C.

Notice that the remaining parts of Theorem 10.2 are either meaninglessor false (compare Theorem 10.2 (vi) with Lemma 11.4 (iii) which shows thatexp : C → C is not injective). We must be very careful in making thetransition from real to complex.

We obtain a series of famous formulae.

4The matter is more subtle than it looks. Classical Euclidean geometry is ‘weaker’than the geometry required to justify analysis and if you wished to obtain analysis fromgeometry you would need to add extra axioms.

23

Lemma 11.4. (i) If θ is real

exp iθ = cos θ + i sin θ.

(ii) If x and y are real

exp(x + iy) = (exp x)(cos y + i sin y).

(iii) exp is periodic with period 2πi, that is to say

exp(z + 2πi) = exp z

for all z ∈ C.(iv) exp(C) = C \ {0}.

Example 11.5. Observe that exp 0 = exp 2πi = 1 but exp′(z) = exp(z) 6= 0for all z ∈ C. Thus the mean value theorem does not hold for differentiablefunctions f : C → C.

It is also possible to extend the definition of sine and cosine to the complexplane but the reader is warned that the behaviour of the new functions maybe somewhat unexpected. Since these extended functions most certainly donot form part of this course (though you will be expected to know them afterthe Complex Methods course) their study is left as an exercise.

Exercise 11.6. (i) Explain why the infinite sums

sin z =∞∑

n=0

(−1)nz2n+1

(2n + 1)!and cos z =

∞∑

n=0

(−1)nz2n

(2n)!

converge everywhere and are differentiable everywhere with sin′ z = cos z,cos′ z = − sin z.

(ii) Show that

sin z =exp iz − exp(−iz)

2i, cos z =

exp iz + exp(−iz))

2

andexp iz = cos z + i sin z

for all z ∈ C.(iii) Show that

sin(z + w) = sin z cos w + cos z sin w, cos(z + w) = cos z cos w − sin z sin w,

24

and (sin z)2 + (cos z)2 = 1 for all z, w ∈ C.(iv) sin(−z) = − sin z, cos(−z) = cos z for all z ∈ C.(v) sin and cos are 2π periodic in the sense that

sin(z + 2π) = sin z and cos(z + 2π) = cos z

for all z ∈ C.(vi) If x is real then sin ix = i sinh x and cos ix = cosh x.(vii) Recover the addition formulae for sinh and cosh by setting z = ix

and w = iy in part (iii).(ix) Show that | sin z| and | cos z| are bounded if |ℑz| ≤ K for some K

but that | sin z| and | cos z| are unbounded on C.

However, as you were already shown in the Algebra and Geometry courseand will be shown again in the Complex Methods course

There is no logarithm defined on all of C \ {0}.

Exercise 11.7. Suppose, if possible, that there exists a continuous L : C \{0} → C with exp(L(z)) = z for all z ∈ C \ {0}.

(i) If θ is real, show that L(exp(iθ)) = i(θ + 2πn(θ)) for some n(θ) ∈ Z.(ii) Define f : R → R by

f(θ) =1

2π

(L(exp iθ) − L(1)

i− θ

)

.

Show that f is a well defined continuous function, that f(θ) ∈ Z for all θ ∈ R,that f(0) = 0 and that f(2π) = −1.

(iii) Show that the statements made in the last sentence of (ii) are in-compatible with the intermediate value theorem and deduce that no functioncan exist with the supposed properties of L.

(iv) Discuss informally what connection, if any, the discussion above haswith the existence of the international date line.

A similar argument shows that it is not possible to produce a continuoussquare root on the complex plane.

Exercise 11.8. Show by modifyiing the argument of Exercise 11.7, that theredoes not exist a continuous S : C → C with S(z)2 = z for all z ∈ C.

More generally, it is not possible to define continuous non-integer powerszα on the complex plane. (Of course, z 7→ zn is well behaved if n is aninteger.) However, in the special case when x is real and strictly positive wecan define

xz = exp(z log x)

25

without problems and this enables us to write exp z = ez where e = exp 1.Surprisingly, Exercise 11.7 is not an end but a beginning of much importantmathematics – but that is another story.

12 The Riemann integral

At school we are taught that an integral is the area under a curve. If pressedthe framer of this definition might reply that everybody knows what area is,but then everybody knows what honey tastes like. But does honey taste thesame to you as it does to me? Perhaps the question is unanswerable but, formany practical purposes, it is sufficient that we agree on what we call honey.

In order to agree on what an integral is, we need a definition which doesnot depend on intuition. It is important that, as far as possible, the propertiesof our formally defined integral shall agree with our intuitive ideas on areabut we have to prove this agreement and not simply assume it.

In this section we introduce a notion of integral due to Riemann. Forthe moment we only attempt to define our integral for bounded functions onbounded intervals.

Let f : [a, b] → R be a function such that there exists a K with |f(x)| ≤ Kfor all x ∈ [a, b]. [To see the connection with ‘the area under the curve’ it ishelpful to suppose initially that 0 ≤ f(x) ≤ K. However, all the definitionsand proofs work more generally for −K ≤ f(x) ≤ K.]

A dissection D of [a, b] is a finite subset of [a, b] containing the end pointsa and b. By convention, we write

D = {x0, x1, . . . , xn} with a = x0 ≤ x1 ≤ x2 ≤ · · · ≤ xn = b.

We define the upper sum and lower sum associated with D by

S(f,D) =n∑

j=1

(xj − xj−1) supx∈[xj−1,xj ]

f(x),

s(f,D) =n∑

j=1

(xj − xj−1) infx∈[xj−1,xj ]

f(x)

[Observe that, if the integral∫ b

af(t) dt exists, then the upper sum ought to

provide an upper bound and the lower sum a lower bound for that integral.]The next lemma is obvious but useful.

Lemma 12.1. If D and D′ are dissections with D′ ⊇ D then

S(f,D) ≥ S(f,D′) ≥ s(f,D′) ≥ s(f,D).

26

The next lemma is again hardly more than an observation but it is thekey to the proper treatment of the integral.

Lemma 12.2. [Key integration property] If f : [a, b] → R is boundedand D1 and D2 are two dissections, then

S(f,D1) ≥ S(f,D1 ∪ D2) ≥ s(f,D1 ∪ D2) ≥ s(f,D2). ⋆

The inequalities ⋆ tell us that, whatever dissection you pick and whateverdissection I pick, your lower sum cannot exceed my upper sum. There is noway we can put a quart in a pint pot5.

Since S(f,D) ≥ −(b − a)K for all dissections D we can define the upperintegral as I∗(f) = infD S(f,D). We define the lower integral similarly asI∗(f) = supD s(f,D). The inequalities ⋆ tell us that these concepts behavewell.

Lemma 12.3. If f : [a, b] → R is bounded, then I∗(f) ≥ I∗(f).

[Observe that, if the integral∫ b

af(t) dt exists, then the upper integral

ought to provide an upper bound and the lower integral a lower bound forthat integral.]

If I∗(f) = I∗(f), we say that f is Riemann integrable and we write∫ b

a

f(x) dx = I∗(f).

The following lemma provides a convenient criterion for Riemann inte-grability.

Lemma 12.4. (i) A bounded function f : [a, b] → R is Riemann integrableif and only if, given any ǫ > 0, we can find a dissection D with

S(f,D) − s(f,D) < ǫ.

(ii) A bounded function f : [a, b] → R is Riemann integrable with integralI if and only if, given any ǫ > 0, we can find a dissection D with

S(f,D) − s(f,D) < ǫ and |S(f,D) − I| ≤ ǫ.

Many students are tempted to use Lemma 12.4 (ii) as the definition of theRiemann integral. The reader should reflect that, without the inequality ⋆,it is not even clear that such a definition gives a unique value for I. (This isonly the first of a series of nasty problems that arise if we attempt to developthe theory without first proving ⋆, so I strongly advise the reader not totake this path.)

We now prove a series of standard results on the integral.

5Or put a litre in a half litre bottle.

27

Lemma 12.5. (i) The function J : [a, b] → R given by J(t) = 1 is integrableand ∫ b

a

1 dx = b − a.

(ii) If f, g : [a, b] → R are Riemann integrable, then so is f + g and

∫ b

a

f(x) + g(x) dx =

∫ b

a

f(x) dx +

∫ b

a

g(x) dx.

(iii) If f : [a, b] → R is Riemann integrable, then −f is and

∫ b

a

(−f(x)) dx = −∫ b

a

f(x) dx.

(iv) If λ ∈ R and f : [a, b] → R is Riemann integrable, then λf isRiemann integrable and

∫ b

a

λf(x) dx = λ

∫ b

a

f(x) dx.

(v) If f, g : [a, b] → R are Riemann integrable functions with f(t) ≥ g(t)for all t ∈ [a, b], then

∫ b

a

f(x) dx ≥∫ b

a

g(x) dx.

Lemma 12.6. (i) If f : [a, b] → R is Riemann integrable then so is f 2.(ii) If f, g : [a, b] → R are Riemann integrable, then so is the product fg.

Lemma 12.7. (i) If f : [a, b] → R is Riemann integrable then so is f+(t) =max(f(t), 0).

(ii) If f : [a, b] → R is Riemann integrable, then |f | is Riemann integrableand ∫ b

a

|f(x)| dx ≥∣∣∣∣

∫ b

a

f(x) dx

∣∣∣∣.

Notice that we often only need the much weaker inequality∣∣∣∣

∫ b

a

f(x) dx

∣∣∣∣≤ sup

t∈[a,b]

|f(t)|(b − a)

usually stated as|integral| ≤ length × sup.

The next lemma is also routine in its proof but continues our programmeof showing that the integral has all the properties we expect.

28

Lemma 12.8. Suppose that a ≤ c ≤ b and that f : [a, b] → R is a boundedfunction. Then, f is Riemann integrable on [a, b] if and only if f |[a,c] is Rie-mann integrable on [a, c] and f |[c,b] is Riemann integrable on [c, b]. Further,if f is Riemann integrable on [a, b], then

∫ b

a

f(x) dx =

∫ c

a

f |[a,c](x) dx +

∫ b

c

f |[c,b](x) dx.

In a very slightly less precise and very much more usual notation we write

∫ b

a

f(x) dx =

∫ c

a

f(x) dx +

∫ b

c

f(x) dx.

There is a standard convention which says that, if b ≥ a and f is Riemannintegrable on [a, b], we define

∫ a

b

f(x) dx = −∫ b

a

f(x) dx.

It is, however, a convention that requires care in use.

Exercise 12.9. Suppose that b ≥ a, λ, µ ∈ R, and f and g are Riemannintegrable. Which of the following statements are always true and which arenot? Give a proof or counter-example. If the statement is not always true,find an appropriate correction and prove it.

(i)

∫ a

b

λf(x) + µg(x) dx = λ

∫ a

b

f(x) dx + µ

∫ a

b

g(x) dx.

(ii) If f(x) ≥ g(x) for all x ∈ [a, b], then

∫ a

b

f(x) dx ≥∫ a

b

g(x) dx.

13 Some properties of the integral

Not all bounded functions are Riemann integrable

Lemma 13.1. If f : [0, 1] → R is given by

f(x) = 1 when x is rational,

f(x) = 0 when x is irrational,

then f is not Riemann integrable

This does not worry us unduly, but makes it more important to showthat the functions we wish to be integrable actually are.

Our first result goes back to Riemann (indeed, essentially, to Newton andLeibniz).

29

Lemma 13.2. (i) If f : [a, b] → R is increasing. then f is Riemann inte-grable.

(ii)If f : [a, b] → R can be written as f = f1 − f2 with f1, f2 : [a, b] → R

increasing, then f is Riemann integrable.(iii) If f : [a, b] → R is piecewise monotonic, then f is Riemann inte-

grable.

It should be noted that the results of Lemma 13.2 do not require f tobe continuous. (For example, the Heaviside function, given by H(t) = 0 fort < 0, H(t) = 1 for t ≥ 0 is increasing but not continuous.) It is quite hardto find a continuous function which is not the difference of two increasingfunctions but an example is given in Exercise 19.18.

The proof of the next result is starred. Next year you will see a simplerproof based on a different idea (that of uniform continuity).

Theorem 13.3. If f : [a, b] → R is continuous then f is Riemann integrable.

We complete the discussion of integration and this course with a series ofresults which apply only to continuous functions.

Our first result is an isolated, but useful, one.

Lemma 13.4. If f : [a, b] → R is continuous, f(t) ≥ 0 for all t ∈ [a, b] and

∫ b

a

f(t) dt = 0,

it follows that f(t) = 0 for all t ∈ [a, b].

Exercise 13.5. Let a ≤ c ≤ b. Give an example of a Riemann integrablefunction f : [a, b] → R such that f(t) ≥ 0 for all t ∈ [a, b] and

∫ b

a

f(t) dt = 0,

but f(c) 6= 0.

We now come to the justly named fundamental theorem of the calculus.

Theorem 13.6. [The fundamental theorem of the calculus] Supposethat f : (a, b) → R is a continuous function and that u ∈ (a, b). If we set

F (t) =

∫ t

u

f(x) dx,

then F is differentiable on (a, b) and F ′(t) = f(t) for all t ∈ (a, b).

30

Exercise 13.7. (i) Let H be the Heaviside function H : R → R given byH(x) = 0 for x < 0, H(x) = 1 for x ≥ 0. Calculate F (t) =

∫ t

0H(x) dx and

show that F is not differentiable at 0. Where does our proof of Theorem 13.6break down?

(ii) Let f(0) = 1, f(t) = 0 otherwise. Calculate F (t) =∫ t

0f(x) dx and

show that F is differentiable at 0 but F ′(0) 6= f(0). Where does our proof ofTheorem 13.6 break down?

Sometimes we think of the fundamental theorem in a slightly differentway.

Theorem 13.8. Suppose that f : (a, b) → R is continuous, that u ∈ (a, b)and c ∈ R. Then there is a unique solution to the differential equationg′(t) = f(t) [t ∈ (a, b)] such that g(u) = c.

We call the solutions of g′(t) = f(t) indefinite integrals (or, simply, inte-grals) of f .

Yet another version of the fundamental theorem is given by the nexttheorem.

Theorem 13.9. Suppose that g : [a, b] → R has continuous derivative. Then

∫ b

a

g′(t) dt = g(b) − g(a).

Theorems 13.6 and 13.9 show that (under appropriate circumstances)integration and differentiation are inverse operations and the the theoriesof differentiation and integration are subsumed in the greater theory of thecalculus.

We use the fundamental theorem of the calculus to prove the formulaefor integration by substitution and integration by parts.

Theorem 13.10. [Change of variables for integrals] Suppose that f :[a, b] → R is continuous and g : [γ, δ] → R is differentiable with continuousderivative. Suppose further that g([γ, δ]) ⊆ [a, b]. Then, if c, d ∈ [γ, δ], wehave ∫ g(d)

g(c)

f(s) ds =

∫ d

c

f(g(x))g′(x) dx.

Exercise 13.11. The following exercise is traditional.(i) Show that integration by substitution, using x = 1/t, gives

∫ b

a

dx

1 + x2=

∫ 1/a

1/b

dt

1 + t2

31

when b > a > 0.(ii) If we set a = −1, b = 1 in the formula of (i), we obtain

∫ 1

−1

dx

1 + x2

?= −

∫ 1

−1

dt

1 + t2

Explain this apparent failure of the method of integration by substitution.(iii) Write the result of (i) in terms of tan−1 and prove it using standard

trigonometric identities.

Lemma 13.12. Suppose that f : [a, b] → R has continuous derivative andg : [a, b] → R is continuous. Let G : [a, b] → R be an indefinite integral of g.Then, we have

∫ b

a

f(x)g(x) dx = [f(x)G(x)]ba −∫ b

a

f ′(x)G(x) dx.

We obtain the following version of Taylor’s theorem by repeated integra-tion by parts.

Theorem 13.13. [A global Taylor’s theorem with integral remain-der] If n ≥ 1 and f : (−a, a) → R is n times differentiable with continuousnth derivative, then

f(t) =n−1∑

j=0

f (j)(0)

j!tj + Rn(f, t)

for all |t| < a, where

Rn(f, t) =1

(n − 1)!

∫ t

0

(t − x)n−1f (n)(x) dx.

In the opinion of the lecturer this form is powerful enough for most pur-poses and is a form that is easily remembered and proved for examination.

14 Infinite integrals

The reader may already be familiar with definitions of the following type.

Definition 14.1. If f : [a, b] → R and M,P ≥ 0 let us write

fM,P (x) =

f(x) if −P ≤ f(x) ≤ M

M if f(x) > M

−P if f(x) < −P .

32

If fM,P is Riemann integrable for each M, P ≥ 0 and∫ b

a

fM,P (x) dx → L

as M,P → ∞ then we say that f is Riemann integrable and∫ b

a

f(x) dx = L

Definition 14.2. If f : [a,∞) → R is such that f |[a,X] is Riemann integrable

for each X > a and∫ X

af(x) dx → L as X → ∞, then we say that

∫∞

af(x) dx

exists with value L.

It must be said that neither Definition 14.1 nor Definition 14.2 are morethan ad hoc.

In the rest of this section we look at Definition 14.2.

Lemma 14.3. Suppose f : [a,∞) → R is such that f |[a,X] is Riemannintegrable on [a,X] for each X > a. If f(x) ≥ 0 for all x, then

∫∞

af(x) dx

exists if and only if there exists a K such that∫ X

af(x) dx ≤ K for all X.

We use Lemma 14.3 to prove the integral comparison test.

Lemma 14.4. If f : [1,∞) → R is a decreasing function with f(x) → 0 asx → ∞ then

∞∑

n=1

f(n) and

∫ ∞

1

f(x) dx

either both diverge or both converge.

Example 14.5. If α > 1 then∑∞

n=1 n−α converges. If α ≤ 1 then∑∞

n=1 n−α

diverges.

This is really as far as we need to go, but I will just add one furtherremark.

Lemma 14.6. Suppose f : [a,∞) → R is such that f |[a,X] is Riemannintegrable on [a,X] for each X > a. If

∫∞

a|f(x)| dx exists, then

∫∞

af(x) dx

exists.

It is natural to state Lemma 14.6 in the form ‘absolute convergence ofthe integral implies convergence’.

Speaking broadly, infinite integrals∫∞

af(x) dx work well when they are

absolutely convergent, that is to say,∫∞

a|f(x)| dx < ∞, but are full of traps

for the unwary otherwise. This is not a weakness of the Riemann integralbut inherent in any mathematical situation where an object only exists ‘byvirtue of the cancellation of two infinite objects’. (Question 19.17 gives anexample of an integral which is convergent but not absolutely convergent.)

33

15 Further reading

The two excellent books Spivak’s Calculus [5] and J. C. Burkill’s A FirstCourse in Mathematical Analysis [2] both cover the course completely andshould be in your college library6. Burkill’s book is more condensed andSpivak’s more leisurely. A completely unsuitable but interesting version ofthe standard analysis course is given by Berlinski’s A Tour of the Calculus [1]— Spivak rewritten by Sterne with additional purple passages by the Ankh-Morpork tourist board. I have written A Companion to Analysis [4] whichcovers this course at a higher level together with the next analysis course.It is available off the web but is unlikely to be as suitable for beginners asSpivak and Burkill. If you do download it, remember that you are under amoral obligation to send me an e-mail about any mistakes you find.

References

[1] D. Berlinski A Tour of the Calculus Mandarin Paperbacks 1997.

[2] J. C. Burkill A First Course in Mathematical Analysis CUP, 1962.

[3] R. P. Burn Numbers and Functions CUP, 1992.

[4] T. W. Korner A Companion to Analysis for the moment available via myhome page http://www.dpmms.cam.ac.uk/˜twk/ .

[5] M. Spivak, Calculus Addison-Wesley/Benjamin-Cummings, 1967.

16 First example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practicein the ideas. Questions and parts of questions marked with ⋆ are not intendedto be hard but cover topics less central to the present course.

Q 16.1. Let an ∈ R. We say that an → ∞ as n → ∞ if, given K > 0, wecan find an n0(K) such that an ≥ K for all n ≥ n0(K).

(i) Write down a similar definition for an → −∞.(ii) Show that an → −∞ if and only if −an → ∞.(iii) If an 6= 0 for all n, show that, if an → ∞, it follows that 1/an → 0

as n → ∞.

6A quieter version of the JCR with shelves of books replacing the bar.

34

(iv) Is it true that, if an 6= 0 for all n and 1/an → 0, then an → ∞ asn → ∞? Give a proof or a counter example.

Q 16.2. Prove that, if

a1 > b1 > 0 and an+1 =an + bn

2, bn+1 =

2anbn

an + bn

,

then an > an+1 > bn+1 > bn. Prove that, as n → ∞, an and bn both tend tothe limit

√(a1b1).

Use this result to give an example of an increasing sequence of rationalnumbers tending to a limit l which is not rational.

Q 16.3. (Exercise 2.2.) Show that a decreasing sequence of real numbersbounded below tends to a limit.[Hint. If a ≤ b then −b ≤ −a.]

Q 16.4. (i) By using the binomial theorem, or otherwise, show that, if η > 0and n is a positive integer, then

(1 + η)n ≥ ηn.

Deduce that (1 + η)n → ∞ as n → ∞.(ii) By using the binomial theorem, or otherwise, show that, if η > 0,

then

(1 + η)n ≥ η2n(n − 1)

2.

Deduce that n−1(1 + η)n → ∞ as n → ∞.(iii) Show that, if k is any positive integer and a > 1, then n−kan → ∞

as n → ∞. [Thus ‘powers beat polynomial growth’.](iv) Show that if k is any positive integer and 1 > a ≥ 0, then nkan → 0

as n → ∞.

Q 16.5. If a ∈ R and a 6= −1 describe the behaviour of

an − 1

an + 1

as n → ∞. (That is, for each value of a say whether the sequence convergesor not, and, if it converges, say what it converges to. For certain values of ayou may find it useful to divide top and bottom by an.)

Q 16.6. (Exercises 3.2 and 3.3.)We work in C.

35

(i) The limit is unique. That is, if an → a and an → b as n → ∞, thena = b.

(ii) If an → a as n → ∞ and n(1) < n(2) < n(3) . . ., then an(j) → a asj → ∞.

(iii) If an = c for all n, then an → c as n → ∞.(iv) If an → a and bn → b as n → ∞, then an + bn → a + b.(v) If an → a and bn → b as n → ∞, then anbn → ab.(vi) If an → a as n → ∞ and an 6= 0 for each n and a 6= 0, then

a−1n → a−1.

(vii) Explain why there is no result in Exercise 3.2 corresponding to part(vii) of Lemma 2.1.

Q 16.7. ⋆[Cesaro summation] If aj ∈ R we write

σn =a1 + a2 + · · · + an

n.

(i) Show that, if an → 0, then σn → 0. [Hint: Given ǫ > 0 we can findan j0(ǫ) such that |aj| < ǫ/2 for all j ≥ j0(ǫ).]

(ii) Show that, if an → a, then σn → a.(iii) If an = (−1)n show that an does not converge but σn does.(iv) If a2m+r = (−1)m for 0 ≤ r ≤ 2m − 1, m ≥ 0 show that σn does not

converge.(v) If σn converges show that n−1an → 0 as n → ∞.

Q 16.8. In this question you may assume the standard properties of theexponential function including the relation expx ≥ 1 + x for x ≥ 0.

(i) Suppose that an ≥ 0. Show that

n∑

j=1

aj ≤n∏

j=1

(1 + aj) ≤ exp

(n∑

j=1

aj

)

.

Deduce, carefully, that∏n

j=1(1 + aj) tends to a limit as n → ∞ if and onlyif∑n

j=1 aj does.(ii)⋆ Euler made use of these ideas as follows. Let pk be the kth prime.

By observing that1

1 − p−1k

≥m∑

r=0

p−rk

show thatn∏

k=1

1

1 − p−1k

≥∑

u∈S(N,n)

1

u

36

where S(N,n) is the set of all integers of the form

u =n∏

k=1

pmk

k

with 0 ≤ mk ≤ N .By letting N → ∞ show that

n∏

k=1

1

1 − p−1k

≥∑

u∈S(n)

1

u

where S(n) is the set of all integers whose only prime factors are p1, p2, . . . ,pn. Deduce that there are infinitely many primes (what can could you sayabout S(n) if there were only n primes?) and show that

n∏

k=1

1

1 − p−1k

→ ∞

as n → ∞.Conclude that

∞∑

k=1

(1

1 − p−1k

− 1

)

diverges.

Show that |1 − (1 − p−1k )−1| ≤ 2p−1

k and deduce that

∞∑

k=1

1

pk

diverges.

Q 16.9. (i) If aj is an integer with 0 ≤ aj ≤ 9 show from the fundamentalaxiom that

∞∑

j=1

aj10−j

exists. Show that 0 ≤∑∞

j=1 aj10−j ≤ 1, carefully quoting any theorems thatyou use.

(ii)⋆ If 0 ≤ x ≤ 1, show that we can find integers xj with 0 ≤ xj ≤ 9 suchthat

x =∞∑

j=1

xj10−j.

(iii)⋆ If aj and bj are integers with 0 ≤ aj, bj ≤ 9 and aj = bj for j < N ,aN > bN show that

∞∑

j=1

aj10−j ≥∞∑

j=1

bj10−j.

Give the precise necessary and sufficient condition for equality and prove it.

37

Q 16.10. ⋆ (We work with the same ideas as in Example 1.1.)(i) Find a differentiable function f : Q → Q such that f ′(x) = 1 for all

x ∈ Q but f(0) > f(1).(ii) If we define g : Q → Q by g(x) = (x2−2)−2 show that g is continuous

but unbounded on the set of x with |x| ≤ 4.

Q 16.11. In each of the following cases determine an integer N (not nec-essarily the least such integer) with the property that if m ≥ N the mthpartial sum of the series

∑∞

n=1 an differs from the sum of the series by lessthan 0.01:

(i) an = 1/n(n + 1);(ii) an = 2−n;(iii)⋆ an = n2−n;(iv)⋆ an = n−n.

Q 16.12. For what values of real β is

∞∑

n=1

nβ

n2β − nβ + 1

convergent and for which divergent. Prove your answer. (You may assumethat

∑∞

n=1 n−β converges if β > 1 and diverges otherwise.)

Q 16.13. ⋆ (i) Let us write

Sn =n∑

r=0

1

r!.

Show by induction, or otherwise, that 1/r! ≤ 2−r+1 for r ≥ 1 and deducethat Sn ≤ 3. Show, from first principles, that Sn converges to a limit (which,with the benefit of extra knowledge, we call e).

(ii) Show that, if n ≥ 2 and r ≥ 0 then

n!

(n + r)!≤ 1

3r.

Deduce carefully that, if m ≥ n ≥ 2,

0 ≤ n!(Sm − Sn) ≤ 1

2.

and that

0 < n!(e − Sn) ≤ 1

2.

Deduce that n!e is not an integer for any n and conclude that e is irrational.

(iii) Show similarly that∞∑

r=0

1

(2r)!is irrational.

38

17 Second example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practicein the ideas. Questions and parts of questions marked with ⋆ are not intendedto be hard but cover topics less central to the present course.

Q 17.1. (Exercises 4.7 and 4.9.)(i)Define the greatest lower bound in the manner of Definition 4.2, prove

its uniqueness in the manner of Lemma 4.3 and state and prove a resultcorresponding to Lemma 4.4.

(ii) Use Lemma 4.8 and Theorem 4.5 to show that any non-empty set inR with a lower bound has a greatest lower bound.

Q 17.2. Suppose that A and B are non-empty bounded subsets of R. Showthat

sup{a + b : a ∈ A, b ∈ B} = sup A + sup B.

The last formula is more frequently written

supa∈A, b∈B

a + b = supa∈A

a + supb∈B

b.

Suppose, further that an and bn are bounded sequences of real numbers.For each of the following statements either give a proof that it is always trueor an example to show that it is sometimes false.

(i) supn(an + bn) = supn an + supn bn.(ii) supn(an + bn) ≤ supn an + supn bn.(iii) supn(an + bn) ≥ supn an + infn bn.(iv) supa∈A, b∈B ab = (supa∈A a)(supb∈B b).(v) infa∈A, b∈B a + b = infa∈A a + infb∈B b.

Q 17.3. (Exercise 5.5) Prove the following results.(i) Suppose that E is a subset of R and that f : E → R is continuous at

x ∈ E. If x ∈ E ′ ⊂ E then the restriction f |E′ of f to E ′ is also continuousat x.

(ii) If J : R → R is defined by J(x) = x for all x ∈ R, then J is continuouson R.

(iii) Every polynomial P is continuous on R.(iv) Suppose that P and Q are polynomials and that Q is never zero on

some subset E of R. Then the rational function P/Q is continuous on E (or,more precisely, the restriction of P/Q to E is continuous.)

39

Q 17.4. (Exercises 5.8 to 5.10.)(i) Show that any real polynomial of odd degree has at least one root. Is

the result true for polynomials of even degree? Give a proof or counterex-ample.

(ii)⋆ Suppose that g : [0, 1] → [0, 1] is a continuous function. By consid-ering f(x) = g(x) − x, or otherwise, show that there exists a c ∈ [0, 1] withg(c) = c. (Thus every continuous map of [0, 1] into itself has a fixed point.)Give an example of a bijective (but, necessarily, non-continuous) functionh : [0, 1] → [0, 1] such that h(x) 6= x for all x ∈ [0, 1].[Hint: First find a function H : [0, 1]\{0, 1, 1/2} → [0, 1]\{0, 1, 1/2} suchthat H(x) 6= x.]

(iii)⋆ Every mid-summer day at six o’clock in the morning, the youngestmonk from the monastery of Damt starts to climb the narrow path up MountDipmes. At six in the evening he reaches the small temple at the peak wherehe spends the night in meditation. At six o’clock in the morning on thefollowing day he starts downwards, arriving back at the monastery at six inthe evening. Of course, he does not always walk at the same speed. Showthat, none the less, there will be some time of day when he will be at thesame place on the path on both his upward and downward journeys.

Q 17.5. (Exercise 6.2.)Let E be a subset of R. Show that function f : E → R is continuous at

x ∈ E if and only if f(y) → f(x) as y → x.

Q 17.6. (Exercise 6.3.)Let E be a subset of R, f , g be some functions from E to R, and x some

point of E. (i) The limit is unique. That is, if f(y) → l and f(y) → k asy → x then l = k.

(ii) If x ∈ E ′ ⊆ E and f(y) → l as y → x through values y ∈ E, thenf(y) → l as y → x through values y ∈ E ′.

(iii) If f(t) = c for all t ∈ E then f(y) → c as y → x.(iv) If f(y) → l and g(y) → k as y → x then f(y) + g(y) → l + k.(v) If f(y) → l and g(y) → k as y → x then f(y)g(y) → lk.(vi) If f(y) → l as y → x, f(t) 6= 0 for each t ∈ E and l 6= 0 then

f(t)−1 → l−1.(vii) If f(t) ≤ L for each t ∈ E and f(y) → l as y → x then l ≤ L.

Q 17.7. (Exercise 6.6.)Let E be a subset of R, f , g be some functions from E to R, and x some

point of E. Prove the following results.(i) If f(t) = c for all t ∈ E then f is differentiable at x with f ′(x) = 0.

40

(ii) If f and g are differentiable at x then so is their sum f + g and

(f + g)′(x) = f ′(x) + g′(x).

(iii) If f and g are differentiable at x then so is their product f × g and

(f × g)′(x) = f ′(x)g(x) + f(x)g′(x).

(iv) If f is differentiable at x and f(t) 6= 0 for all t ∈ E then 1/f isdifferentiable at x and

(1/f)′(x) = −f ′(x)/f(x)2.

(v) If f(t) =∑N

r=0 artr on E then then f is differentiable at x and

f ′(x) =N∑

r=1

rarxr−1

.

Q 17.8. We work in the real numbers. Are the following true or false? Givea proof or counterexample as appropriate.

(i) If∑∞

n=1 a4n converges then

∑∞

n=1 a5n converges.

(ii) If∑∞

n=1 a5n converges then

∑∞

n=1 a4n does.

(iii) If an ≥ 0 for all n and∑∞

n=1 an converges then nan → 0 as n → ∞.(iv)⋆ If an ≥ 0 for all n and

∑∞

n=1 an converges then n(an − an−1) → 0 asn → ∞.

(v)⋆ If an is a decreasing sequence of positive numbers and∑∞

n=1 an con-verges then nan → 0 as n → ∞.

(vi)⋆ If an is a decreasing sequence of positive numbers and nan → 0 asn → ∞ then

∑∞

n=1 an converges.[Hint. If necessary, look at Lemmas 14.4 and 14.5.]

(vii)⋆ If∑∞

n=1 a2n converges then

∑∞

n=1 n−1an converges.[Hint: Cauchy-Schwarz]

(viii)⋆ If∑∞

n=1 an converges then∑∞

n=1 n−1|an| converges.

Q 17.9. ⋆ [General principle of convergence] We say that a sequence xn

of points in R form a Cauchy sequence if given any ǫ > 0 we can find ann0(ǫ) such that

|xn − xm| < ǫ for all n, m ≥ n0(ǫ).

(i) Show that any convergent sequence is a Cauchy sequence.(ii) Suppose that the points xn form a Cauchy sequence. Show that, if

we can find n(j) → ∞ such that xn(j) → x, it follows that xn → x. (Thus, ifany subsequence of a Cauchy sequence converges, so does the sequence.)

41

(iii) Show that if the points xn form a Cauchy sequence the set {xn : n ≥1} is bounded.

(iv) Use the theorem of Bolzano-Weierstrass to show that any Cauchysequence converges.

[We have thus shown that a sequence converges if and only if it is a Cauchysequence. This is the famous Cauchy general principle of convergence. Itsgeneralisation will play an important role in next year’s analysis course C9.]

Q 17.10. ⋆ [A method of Abel](i) Suppose that aj and bj are sequences of complex numbers and that

Sn =∑n

j=1 aj for n ≥ 1 and S0 = 0. Show that, if 1 ≤ u ≤ v then

v∑

j=u

ajbj =v∑

j=u

Sj(bj − bj+1) − Su−1bu + Svbv+1.

(This is known as partial summation, for obvious reasons.)(ii) Suppose now that, in addition, the bj form a decreasing sequence of

positive terms and that |Sn| ≤ K for all n. Show that∣∣∣∣∣

v∑

j=u

ajbj

∣∣∣∣∣≤ 3Kbu.

(You can replace 3Kbu by 2Kbu if you are careful but there is no advantagein this.) Deduce that if bj → 0 as j → ∞ then

∑∞

j=1 ajbj converges.Deduce the alternating series test.(iii) If bj is a decreasing sequence of positive terms with bj → 0 as j → ∞

show that∑∞

j=1 bjzj converges in the region given by |z| ≤ 1 and z 6= 1.

Show by example that we must have the condition z 6= 1. Show by examplethat we must have the condition |z| ≤ 1.

Q 17.11.⋆ Enter any number on your calculator. Press the sin button repeat-edly. What appears to happen? Prove your conjecture using any propertiesof sin that you need.

18 Third example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practicein the ideas. Questions and parts of questions marked with ⋆ are not intendedto be hard but cover topics less central to the present course. I know thatthere is a general belief amongst students, directors of studies and the Faculty

42

Board that there is a magic set of questions which is suitable for everybody.If there is one, I will be happy to circulate it. For the moment I remark thatthe unstarred questions on this example sheet represent a good week’s workfor someone who finds the course hard and the whole sheet represents a goodweek’s work for someone who finds the unstarred questions boring.

Q 18.1. [Very traditional] (i) Consider the function f : R → R given byf(t) = t2 sin 1/t for t 6= 0, f(0) = 0. Show that f is everywhere differentiableand find its derivative. Show that f ′ is not continuous.[Deal quickly with the easy part and then go back to the definition to dealwith t = 0. There are wide selection of counter-examples obtained by lookingat tβ sin tα for various values of α and β.]

(ii) Find an infinitely differentiable function g : (1,∞) → R such thatg(t) → 0 but g′(t) 9 0 as t → ∞.

Q 18.2.⋆ Question 18.1 shows that the derivative of a differentiable functionneed not be continuous. In spite of this the derivative still obeys Darboux’stheorem:- If f : [a, b] → R is differentiable and k lies between f ′(a) and f ′(b)then there is a c ∈ [a, b] such that f ′(c) = k. In this question we prove theresult.

Explain why there is no loss of generality in supposing that f ′(a) > k >f ′(b). Set g(x) = f(x)−kx. By looking at g′(a) and g′(b) show that g cannothave a maximum at a or b. Use the method of proof of Rolle’s theorem toshow that there exists a c ∈ (a, b) with g′(c) = 0 and deduce Darboux’stheorem.

Give an example of a function f : R → R such that there does not exista differentiable function F : R → R with F ′ = f .

Q 18.3. (i) [Cauchy’s mean value theorem] Suppose that f, g : [a, b] →R are continuous and that f and g are differentiable on (a, b). Supposefurther that g′(x) 6= 0 for all x ∈ (a, b). Explain why g(a) 6= g(b). Byapplying Rolle’s theorem to F where

F (x) = (g(b) − g(a))(f(x) − f(a)) − (g(x) − g(a))(f(b) − f(a)),

show that there is a ζ ∈ (a, b) such that

f ′(ζ)

g′(ζ)=

f(b) − f(a)

g(b) − g(a).

(ii) [L’Hopital’s rule] Suppose that f, g : [a, b] → R are continuous andthat f and g are differentiable on (a, b). Suppose that f(a) = g(a) = 0, thatg′(t) does not vanish near a and f ′(t)/g′(t) → l as t → a through values oft > a. Show that f(t)/g(t) → l as t → a through values of t > a.

43

Q 18.4. (Exercise 7.9.)Consider the function F : R → R defined by

F (0) = 0

F (x) = exp(−1/x2) otherwise.

(i) Prove by induction, using the standard rules of differentiation, that Fis infinitely differentiable at all points x 6= 0 and that, at these points,

F (n)(x) = Pn(1/x) exp(−1/x2)

where Pn is a polynomial which need not be found explicitly.(ii) Explain why x−1Pn(1/x) exp(−1/x2) → 0 as x → 0.(iii) Show by induction, using the definition of differentiation, that F is

infinitely differentiable at 0 with F (n)(0) = 0 for all n. [Be careful to get thispart of the argument right.]

(iv) Show that

F (x) =∞∑

j=0

F (j)(0)

j!xj

if and only if x = 0. (The reader may prefer to say that ‘The Taylor expansionof F is only valid at 0’.)

(v) Why does part (iv) not contradict the local Taylor theorem (Theo-rem 7.8)?

Q 18.5. In this question you may assume the standard properties of sin andcos but not their power series expansion.

(i) By considering the sign of f ′1(x), when f1(t) = t − sin t, show that

t ≥ sin t

for all t ≥ 0.(ii) By considering the sign of f ′

2(x), when f2(t) = cos t− 1 + t2/2!, showthat

cos t ≥ 1 − t2

2!

for all t ≥ 0.(iii) By considering the sign of f ′

3(x), when f3(t) = sin t− t + t3/3!, showthat

sin t ≥ t − t3

3!

for all t ≥ 0.

44

(iv) State general results suggested by parts (i) to (iii) and prove themby induction. State and prove corresponding results for t < 0.

(v) Using (iv), show that

N∑

n=0

(−1)nt2n+1

(2n + 1)!→ sin t

as N → ∞ for all t ∈ R. State and prove a corresponding result for cos.[This question could be usefully illustrated by computer graphics.]

Q 18.6. (i) Suppose f : [0, 1] → R is twice differentiable with f ′′(t) ≥ 0 forall t ∈ [0, 1]. If f ′(0) > 0 and f(0) = 0 explain why f(t) > 0 for t > 0.If f ′(0) ≥ 0 and f(0) = f(1) = 0 what can you say about f and why? Iff ′(1) ≤ 0 and f(0) = f(1) = 0 what can you say about f and why?

(ii) Suppose f : [0, 1] → R is twice differentiable with f ′′(t) ≥ 0 for allt ∈ [0, 1] and f(0) = f(1) = 0. Show that f(t) ≤ 0 for all t ∈ [0, 1]. [Hint:Consider the sign of f ′.]

(iii) Suppose g : [a, b] → R is twice differentiable with g′′(t) ≥ 0 for allt ∈ [a, b]. By considering the function f : [0, 1] → R defined by

f(t) = g((1 − t)a + tb

)− (1 − t)g(a) − tg(b)

show thatg((1 − t)a + tb

)≤ (1 − t)g(a) + tg(b)

for all t with 1 ≥ t ≥ 0.[In other words a twice differentiable function with everywhere positive

second derivative is convex. Convex functions are considered in the lastpart of the probability course where you prove the very elegant Jensen’sinequality. You should note, however, that not all convex functions are twicedifferentiable (look at x 7→ |x|).]

Q 18.7.⋆ The results of this question are also useful in the probability coursewhen you study extinction probabilities. Notice that the point of this ques-tion is to obtain rigorous proofs of the results stated.

Suppose a0 > 0, a1, a2, . . . , an ≥ 0 and∑n

j=0 aj = 1. We set P (t) =∑n

j=0 ajtj.

(i) Find P (0), P (1) and P ′(1). Show that P ′(t) ≥ 0 and P ′′(t) ≥ 0 for allt ≥ 0.

(ii) Show that the equation P (t) = t has no solution with 0 ≤ t < 1 if∑n

j=1 jaj ≤ 1 and exactly one such solution if∑n

j=1 jaj > 1. We write α forthe smallest solution of P (t) = t with 0 ≤ t ≤ 1.

45

(iii) If we set x0 = 0 and xn = P (xn−1) for n ≥ 1 show, by induction,that xn−1 ≤ xn ≤ α.

(iv) Deduce, giving your reasons explicitly, that xn must converge to alimit β. Show that 0 ≤ β ≤ α and P (β) = β. Deduce that β = α and so

xn → α as n → ∞.

Q 18.8.⋆ The first proof that transcendental numbers existed is due to Li-ouville. This question gives a version of his proof.

Let P (t) =∑N

j=0 ajtj with the aj integers and aN 6= 0. Let α be a root

of P .(i) Why can we choose a δ > 0 such that P (t) 6= 0 for all t with t ∈

[α − δ, α + δ] and t 6= α.(ii) Let δ be as in (i). Explain why there exists a K such that |P (t) −

P (s)| ≤ K|t − s| for all t, s ∈ [α − δ, α + δ].(iii) If p and q are integers such that q ≥ 1 and P (p/q) 6= 0 show that

|P (p/q)| ≥ q−N . [Hint: Remember that the coefficients of P are integers.](iv) If p and q are integers such that q ≥ 1, P (p/q) 6= 0 and p/q ∈

[α − δ, α + δ] use (ii) and (iii) to show that

∣∣∣∣α − p

q

∣∣∣∣≥ K−1q−N .

(v) Show that, if p and q are integers such that q ≥ 1 and α 6= p/q, then

∣∣∣∣α − p

q

∣∣∣∣≥ min(δ,K−1q−N).

(vi) Show that if α is a root of a polynomial with integer coefficients thenthere exists a K and an N (depending on α) such that

∣∣∣∣α − p

q

∣∣∣∣≥ K−1q−N

whenever p and q are integers with q ≥ 1 and α 6= p/q.(vii) Explain why

∑∞

n=0 10−n! converges. Call the limit L. Show that

∣∣∣∣∣L −

m∑

n=0

10−n!

∣∣∣∣∣≤ 2((10)m!)−m−1.

By taking q = (10)m! and looking at (vi), show that L can not be the rootof a polynomial with integer coefficients.

46

(viii) By looking at∞∑

n=0

ζn10−n!

with ζn = ±1 show that there are uncountably many transcendentals.

Q 18.9. In this question you may assume standard results on the powerfunction t 7→ tα.

Use the form of Taylor’s theorem given in Theorem 7.7 to show that

(1 + x)α = 1 + αx +α(α − 1)

2!x2 + · · · + α(α − 1) . . . (α − n + 1)

n!xn + . . .

for all 0 ≤ x < 1. Note that the main part of your task is to estimate theremainder term.

What happens if α is a positive integer?Why does your argument fail for −1 < x < 0? (The result is true but

our form of the remainder does not give it. This is one of the reasons whythere are so many forms of Taylor’s with different remainders.)

Q 18.10. ⋆ (i) Suppose that x0, x1, . . .xn are distinct. Set

ej(x) =∏

k 6=j

x − xk

xj − xk

.

What is the value of ej(xk) if j 6= k and if j = k? Show that given realnumbers αj there is a polynomial of degree n with f(xj) = αj.

(ii) Suppose f : [a, b] → R is n + 1 times differentiable, a < x0 < x1 <· · · < xn < b and P is a polynomial of degree n with P (xj) = f(xj) for0 ≤ j ≤ n. We are interested in the error

e(x) = f(x) − P (x)

at a point x ∈ [a, b] with x 6= xj for all j.Set

F (t) = f(t) − P (t) − e(x)n∏

k=0

t − xk

x − xk

.

Explain why F vanishes at least n+2 times on [a, b]. Explain why F ′ vanishesat least n + 1 times on [a, b]. By repeating the argument show that F (n+1)

must vanish at least once (at ζ, say) on [a, b]. Show that

e(x) =1

(n + 1)!f (n+1)(ζ)

n∏

k=0

(x − xk).

47

Deduce that if |f (n+1)(t)| ≤ M for all t ∈ [a, b] then

|f(t) − P (t)| ≤ M

(n + 1)!|Q(t)|

where Q(t) =∏n

k=0(t − xk).[The argument just given should remind you of the proof of Theorem 7.6.]

Q 18.11. ⋆ Recall from last term (or earlier) that

(cos θ + i sin θ)n = cos nθ + i sin nθ

for all real θ and all integers n ≥ 0.By taking real parts, show that there is a real polynomial Tn of degree n

such thatTn(cos θ) = cos nθ

for all real θ.Write down T0(t), T1(t), T2(t) and T3(t) explicitly.Show that, if n ≥ 1 then(a) |Tn(t)| ≤ 1 for all t with |t| ≤ 1,(b) Tn+1(t) = 2tTn(t) − Tn−1(t), (why does this result hold for all t?)(c) the coefficient of tn in Tn(t) is 2n−1.Explain why Tn+1 has n+1 zeros x0, x1, . . . , xn lying in (−1, 1). Use (a),

(c) and the final result of question 18.10 to show that if [a, b] = [−1, 1]and f and P obey the hypotheses of question 18.10 (so that, in particular,|f (n+1)(t)| ≤ M for all t ∈ [−1, 1]) then

|f(t) − P (t)| ≤ M

(n + 1)!|2−nTn+1(t)| ≤

M

2n(n + 1)!

The rest of the question just asks for another of useful property of theTchebychev polynomials Tn. (The modern view is that Tchebychev shouldhave called himself Chebychev. He seems to have preferred Tchebycheff.)

(d) If n, m ≥ 0 then

∫ 1

−1

Tn(x)Tm(x)

(1 − x2)1/2dx =

{

0 if n 6= mπ2

if n = m 6= 0

What happens if n = m = 0?

Q 18.12. (Exercise 8.4.)Let E be a subset of C, f , g be some functions from E to C, and z some

point of E.

48

(i) The limit is unique. That is, if f(w) → l and f(w) → k as w → zthen l = k.

(ii) If z ∈ E ′ ⊆ E and f(w) → l as w → z through values w ∈ E, thenf(w) → l as w → z through values w ∈ E ′.

(iii) If f(u) = c for all u ∈ E then f(w) → c as w → z.(iv) If f(w) → l and g(w) → k as w → z then f(w) + g(w) → l + k.(v) If f(w) → l and g(w) → k as w → z then f(w)g(w) → lk.(vi) If f(w) → l as w → z, f(u) 6= 0 for each u ∈ E and l 6= 0 then

f(w)−1 → l−1.

Q 18.13. (Exercise 8.5 and 8.6.) Suppose that E is a subset of C, thatz ∈ E, and that f and g are functions from E to C.

(i) If f(u) = c for all u ∈ E, then f is continuous on E.(ii) If f and g are continuous at z, then so is f + g.(iii) Let us define f × g : E → C by f × g(u) = f(u)g(u) for all u ∈ E.

Then if f and g are continuous at z, so is f × g.(iv) Suppose that f(u) 6= 0 for all u ∈ E. If f is continuous at z so is

1/f .(v) If z ∈ E ′ ⊂ E and f is continuous at z then the restriction f |E′ of f

to E ′ is also continuous at z.(vi) If J : C → C is defined by J(z) = z for all z ∈ C, then J is continuous

on C.(vii) Every polynomial P is continuous on C.(viii) Suppose that P and Q are polynomials and that Q is never zero on

some subset E of C. Then the rational function P/Q is continuous on E (or,more precisely, the restriction of P/Q to E is continuous.)

(ix) Let U and V be subsets of C. Suppose f : U → C is such thatf(z) ∈ V for all z ∈ U . If f is continuous at w ∈ U and g : V → C iscontinuous at f(w), then the composition g ◦ f is continuous at w.

Q 18.14. (Exercises 8.8 8.9 and 8.10.)Let E be a subset of C, f , g be some functions from E to C, and z some

point of E.Show that if f is differentiable at z then f is continuous at z.Prove the following results.(i) If f(u) = c for all u ∈ E then f is differentiable at z with f ′(z) = 0.(ii) If f and g are differentiable at z then so is their sum f + g and

(f + g)′(z) = f ′(z) + g′(z).

(iii) If f and g are differentiable at z then so is their product f × g and

(f × g)′(z) = f ′(z)g(z) + f(z)g′(z).

49

(iv) If f is differentiable at z and f(u) 6= 0 for all u ∈ E then 1/f isdifferentiable at z and

(1/f)′(z) = −f ′(z)/f(z)2.

(v) If f(u) =∑N

r=0 arur on E then then f is differentiable at z and

f ′(z) =N∑

r=1

rarzr−1

.(vi) Let U and V be subsets of C. Suppose f : U → C is such that

f(z) ∈ V for all t ∈ U . If f is differentiable at w ∈ U and g : V → R isdifferentiable at f(w), then the composition g ◦ f is differentiable at w with

(g ◦ f)′(w) = f ′(w)g′(f(w)).

Q 18.15. (Exercise 9.3.) Suppose that∑∞

n=0 anzn has radius of convergence

R. Then the sequence |anzn| is unbounded if |z| > R and

∑∞

n=0 anzn con-

verges absolutely if |z| < R.

Q 18.16.⋆ This question requires Abel’s test from Exercise 17.10 which isalso starred.

(i) Show that∑∞

n=1 zn/n has radius of convergence 1, converges for all zwith |z| = 1 and z 6= 1 but diverges if z = 1.

(ii) Let |z1| = |z2| = · · · = |zm| = 1. Find a power series∑∞

n=0 anzn

which has radius of convergence 1 and converges for all z with |z| = 1 andz 6= z1, z2, . . . , zm but diverges if z = zj for some 1 ≤ j ≤ m.

Q 18.17. ⋆ (i) Show that there exist power series of all radii of convergence(including 0 and ∞).

(ii) Suppose the power series∑∞

n=0 anzn has radius of convergence R andthe power series

∑∞

n=0 bnzn has radius of convergence S. Show that, if R 6= S

then∑∞

n=0(an + bn)zn has radius of convergence min(R,S).(iii) Suppose that the conditions of (ii) hold except that R = S. Show

that the radius of convergence T of∑∞

n=0(an + bn)zn satisfies the conditionT ≥ R. Show by means of examples that T can take any value with T ≥ R.[Hint: Start by thinking of a simple relation between an and bn which willgive T = ∞.]

Q 18.18.⋆ (i) Suppose that xn is a bounded sequence of real numbers. Showcarefully that yn = supm≥n xm is a bounded decreasing sequence and deducethat yn tends to a limit y say. We write

lim supn→∞

xn = y.

50

(ii) Suppose that xn is a bounded sequence of real numbers. Prove thefollowing two results.

(A) Given any ǫ > 0, we can find an M(ǫ) such that

xn ≤ lim supm→∞

xm + ǫ

for all n ≥ M(ǫ).(B) Given any ǫ > 0 and any N we can find an integer P (N, ǫ) ≥ N

such thatxP (N,ǫ) ≥ lim sup

m→∞

xm − ǫ.

(iii) Using (ii) or otherwise show that if an ∈ C and the sequence |an|1/n

is bounded and lim supn→∞ |an|1/n 6= 0 then∑∞

n=0 anzn has radius of conver-

gence (lim supn→∞ |an|1/n)−1.Show also that, if the sequence |an|1/n is unbounded,

∑∞

n=0 anzn has ra-

dius of convergence 0, and, if lim supn→∞ |an|1/n = 0,∑∞

n=0 anzn has infinite

radius of convergence.[This result has considerable theoretical significance but I strongly advise

using the definition directly rather than relying on the formula.](iv) Use the formula of this question to obtain the results of Ques-

tion 18.17.

Q 18.19. (i) Starting from the observation that

1 + x + x2 + · · · + xn =1 − xn+1

1 − x

show that1

1 − x=

∞∑

n=0

xn

for all |x| < 1. Show that the result is false or meaningless if |x| ≥ 1.(ii) Use term by term differentiation to obtain power series expansions

for (1 − x)−n for all integer n with n ≥ 1.(iii) Find the radius of convergence of

∑∞

n=1 xn/n. Use term by termdifferentiation and the constant value theorem to show that

log(1 − x) = −∞∑

n=1

xn

n.

(iv) Use similar ideas to obtain a power series expansion for tan−1 x in arange to be stated.

51

Q 18.20. (Exercise 10.10.)Write down what you consider to be the chief properties of sinh and cosh.

(They should convey enough information to draw a reasonable graph of thetwo functions.)

(i) Obtain those properties by starting with the definitions

sinh x =∞∑

n=0

x2n+1

(2n + 1)!and cosh x =

∞∑

n=0

x2n

(2n)!

and proceeding along the lines of Lemma 10.9.(ii) Obtain those properties by starting with the definitions

sinh x =exp x − exp(−x)

2and cosh x =

exp x + exp(−x)

2

Q 18.21. ⋆(Exercise 11.6.)(i) Explain why the infinite sums

sin z =∞∑

n=0

(−1)nz2n+1

(2n + 1)!and cos z =

∞∑

n=0

(−1)nz2n

(2n)!

converge everywhere and are differentiable everywhere with sin′ z = cos z,cos′ z = − sin z.

(ii) Show that

sin z =exp iz − exp(−iz)

2i, cos z =

exp iz + exp(−iz)

2

andexp iz = cos z + i sin z

for all z ∈ C.(iii) Show that

sin(z + w) = sin z cos w + cos z sin w, cos(z + w) = cos z cos w − sin z sin w,

and (sin z)2 + (cos z)2 = 1 for all z, w ∈ C.(iv) sin(−z) = − sin z, cos(−z) = cos z for all z ∈ C.(v) sin and cos are 2π periodic in the sense that

sin(z + 2π) = sin z and cos(z + 2π) = cos z

for all z ∈ C.(vi) If x is real then sin ix = i sinh x and cos ix = cosh x.(vii) Recover the addition formulae for sinh and cosh by setting z = ix

and w = iy in part (iii).(viii) Show that | sin z| and | cos z| are bounded if |ℑz| ≤ K for some K

but that | sin z| and | cos z| are unbounded on C.

52

Q 18.22. ⋆(Exercise 11.7.)Suppose, if possible, that there exists a continuous L : C \ {0} → C with

exp(L(z)) = z for all z ∈ C \ {0}.(i) If θ is real, show that L(exp(iθ)) = i(θ + 2πn(θ)) for some n(θ) ∈ Z.(ii) Define f : R → R by

f(θ) =1

2π

(L(exp iθ) − L(1)

i− θ

)

.

Show that f is a well defined continuous function, that f(θ) ∈ Z for all θ ∈ R,that f(0) = 0 and that f(2π) = −1.

(iii) Show that the statements made in the last sentence of (ii) are incom-patible with the intermediate value theorem and deduce that no function canexist with the supposed properties of L.

(iv) Discuss informally what connection, if any, the discussion above haswith the existence of the international date line.

Q 18.23.⋆(Exercise 11.8.) Show by modifying the argument of Exercise 11.7,that there does not exist a continuous S : C → C with S(z)2 = z for all z ∈ C.

19 Fourth example sheet

Students vary in how much work they are prepared to do. On the whole,exercises from the main text are reasonably easy and provide good practicein the ideas. Questions and parts of questions marked with ⋆ are not intendedto be hard but cover topics less central to the present course.

Q 19.1. (Exercise 12.9.) Suppose that b ≥ a, λ, µ ∈ R, and f and g areRiemann integrable. Which of the following statements are always true andwhich are not? Give a proof or counter-example. If the statement is notalways true, find an appropriate correction and prove it.

(i)

∫ a

b

λf(x) + µg(x) dx = λ

∫ a

b

f(x) dx + µ

∫ a

b

g(x) dx.

(ii) If f(x) ≥ g(x) for all x ∈ [a, b], then

∫ a

b

f(x) dx ≥∫ a

b

g(x) dx.

Q 19.2. (Exercise 13.5.) Let a ≤ c ≤ b. Give an example of a Riemannintegrable function f : [a, b] → R such that f(t) ≥ 0 for all t ∈ [a, b] and

∫ b

a

f(t) dt = 0,

but f(c) 6= 0.

53

Q 19.3.⋆ Define f : [0, 1] → R by f(p/q) = 1/q when p and q are coprimeintegers with 1 ≤ p < q and f(x) = 0 otherwise.

(i) Show that f is Riemann integrable and find∫ 1

0f(x) dx.

(ii) At which points is f continuous? Prove your answer.

Q 19.4. (Exercise 13.7).(i) Let H be the Heaviside function H : R → R given by H(x) = 0 for

x < 0, H(x) = 1 for x ≥ 0. Calculate F (t) =∫ t

0H(x) dx and show that F is

not differentiable at 0. Where does our proof of Theorem 13.6 break down?(ii) Let f(0) = 1, f(t) = 0 otherwise. Calculate F (t) =

∫ t

0f(x) dx and

show that F is differentiable at 0 but F ′(0) 6= f(0). Where does our proof ofTheorem 13.6 break down?

Q 19.5. (Exercise 13.11.)The following exercise is traditional.(i) Show that integration by substitution, using x = 1/t, gives

∫ b

a

dx

1 + x2=

∫ 1/a

1/b

dt

1 + t2

when b > a > 0.(ii) If we set a = −1, b = 1 in the formula of (i), we obtain

∫ 1

−1

dx

1 + x2

?= −

∫ 1

−1

dt

1 + t2

Explain this apparent failure of the method of integration by substitution.(iii) Write the result of (i) in terms of tan−1 and prove it using standard

trigonometric identities.

Q 19.6. In this question we give an alternative treatment of the logarithmso no properties of the exponential or logarithmic function should be used.You should quote all the theorems that you use, paying particular attentionto those on integration.

We set

l(x) =

∫ x

1

1

tdt.

(i) Explain why l : (0,∞) → R is a well defined function.(ii) Use the change of variable theorem for integrals to show that

∫ xy

x

1

tdt = l(y)

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for all x, y > 0. Deduce that l(xy) = l(x) + l(y).(iii) Show that l is everywhere differentiable with l′(x) = 1/x.(iv) Show that l is a strictly increasing function.(v) Show that l(x) → ∞ as x → ∞.

Q 19.7. In the lectures we deduced the properties of the logarithm fromthose of the exponential. Reverse this by making a list of the properties ofthe exponential, define exp as the inverse function of log (explaining carefullywhy you can do this) and using the properties of log found in the previousquestion (Question 19.6).

Q 19.8.⋆ [The first mean value theorem for integrals] Suppose g :[a, b] → R is a continuous function such that g(x) ≥ 0 for all x ∈ [a, b]. Iff : [a, b] → R is continuous explain why we can find k1 and k2 in [a, b] suchthat

f(k1) ≤ f(x) ≤ f(k2)

for all x ∈ [a, b]. Deduce carefully that

f(k1)

∫ b

a

g(x) dx ≤∫ b

a

f(x)g(x) dx ≤ f(k2)

∫ b

a

g(x) dx

and show, stating carefully any theorem that you need, that there exists ac ∈ [a, b] such that

∫ b

a

f(x)g(x) dx = f(c)

∫ b

a

g(x) dx.

Does this result remain true if g(x) ≤ 0 for all x ∈ [a, b]? Does this resultremain true if we place no restrictions on g (apart from continuity). In eachcase give a proof or a counter example. (The first mean value theorem forintegrals is used in the numerical analysis course.)

Q 19.9. Use the integral form of the remainder in Taylor’s theorem (i.e.Theorem 13.13) to obtain the power series expansion for sin.

Q 19.10. [The binomial theorem] If 1 > x ≥ t ≥ 0 show that

x − t

1 + t≤ x.

If −1 < x ≤ t ≤ 0 show that

t − x

1 + t≤ −x.

Use the integral form of the Taylor theorem to show that, if |x| < 1, then

(1 + x)α = 1 + αx +α(α − 1)

2!x2 + · · · + α(α − 1) . . . (α − n + 1)

n!ixn + . . . .

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Q 19.11.⋆ [Cauchy-Schwarz for integrals] Write C([a, b]) for the set ofcontinuous functions f : [a, b] → R.

(i) If you are doing course P1 verify that C([a, b]) is a vector space overR. Is it finite dimensional? Prove your answer.

(ii) If f, g ∈ C([a, b]) we write

〈f, g〉 =

∫ b

a

f(t)g(t) dt.

Show that if f, g, h ∈ C([a, b]) and λ, µ ∈ R then(a) 〈f, f〉 ≥ 0,(b) if 〈f, f〉 = 0 then f = 0.(c) 〈f, g〉 = 〈g, f〉,(d) 〈λf + µg, h〉 = λ〈f, h〉 + µ〈g, h〉.

(iii) By imitating the proof of the Cauchy-Schwarz inequality for Rn inthe course C1/C2 show that

〈f, g〉2 ≤ 〈f, f〉〈g, g〉.

In other words

(∫ b

a

f(t)g(t) dt

)2

≤∫ b

a

f(t)2 dt

∫ b

a

g(t)2 dt.

(iv) When do we have equality in the inequality proved in (iii).

Q 19.12. ⋆ Explain why

1

n + 1≤∫ n+1

n

1

xdx ≤ 1

n.

Hence or otherwise show that if we write

Tn =n∑

r=1

1

r− log n

we have Tn+1 ≤ Tn for all n ≥ 1. Show also that 1 ≥ Tn ≥ 0. Deduce thatTn tends to a limit γ (Euler’s constant) with 1 ≥ γ ≥ 0. [It is an indicationof how little we know about specific real numbers that, after three centuries,we still do not know whether γ is irrational. G. H. Hardy is said to haveoffered his chair to anyone who could prove that γ was transcendental.]

(ii) By considering T2n − Tn, show that

1 − 1

2+

1

3− 1

4+

1

5− 1

6+

1

7− 1

8· · · = log 2

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(iii) By considering T4n − 12T2n − 1

2Tn, show that

1 +1

3− 1

2+

1

5+

1

7− 1

4· · · =

3

2log 2

The famous example is due to Dirichlet. It gives a specific example whererearranging a non-absolutely convergent sum changes its value.

Q 19.13. [Simple versions of Stirling’s formula] (The first part of thisquestion is in the probability course.)

(i) Prove that∫ n

1

log x dx ≤n∑

r=1

log r ≤∫ n

1

log x dx + log n.

Compute∫ n

1log x dx and hence show that

1

n(n!)1/n → 1

eas n → ∞.

(ii)⋆ Show that∫ n+1/2

n−1/2

log x dx − log n =

∫ 1/2

0

log

(

1 +t

n

)

+ log

(

1 − t

n

)

dt.

By using the mean value theorem, or otherwise, deduce that∣∣∣∣∣

∫ n+1/2

n−1/2

log x dx − log n

∣∣∣∣∣≤ 4

3n2.

(You may replace 4/(3n2) by An−2 with A another constant, if you wish.)

Deduce that∫ N+1/2

1/2log x dx − log N ! converges to a limit. Conclude that

n!

(n + 1/2)(n+1/2)en → C

as n → ∞ for some constant C.(iii)⋆ Find a function f(n) not involving factorials such that

f(n)

(2n

n

)

tends to a limit as n → ∞. You should try for an f in as simple a form aspossible. (But, of course, different people may have different views on whatsimple means.)

(iv)⋆ Show that the result of (ii) implies the result of (i). Give an example

of a sequence an such that a1/nn converges but an does not.

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Q 19.14. (i) Show that∑∞

n=271

n log n log log ndiverges. Find a value of N

such that∑N

n=271

n log n log log n≥ 3. Try and find its numerical value on your

calculator.(ii)∗ Given an > 0 such that

∑∞

n=1 an diverges show that we can findbn > 0 such that bn/an → 0 but

∑∞

n=1 bn diverges.Given an > 0 such that

∑∞

n=1 an converges show that we can find bn > 0such that bn/an → ∞ but

∑∞

n=1 bn converges.[These two results show that it is futile to look for some sort of ‘super-

charged ratio test’ to decide the convergence of all possible series.]

Q 19.15. (i) By writing

r(r + 1) . . . (r + m− 1) = Am

(r(r + 1) . . . (r + m)− (r − 1)r . . . (r + m− 1)

),

where Am is to found explicitly, compute∑N

r=1 r(r+1) . . . (r+m−1). Deducethat

N−m−1

N∑

r=1

r(r + 1) . . . (r + m − 1) → 1

m + 1.

(ii) Show that

r(r + 1) . . . (r + m − 1) − rm = P (r)

where P is a polynomial of degree less than m. Show using (i) and induction,or otherwise, that

N−m−1

N∑

r=1

rm → 1

m + 1.

(iii) Use dissections of the form

D = {0, a/n, 2a/n, . . . , a}

to compute ∫ a

0

xm dx

directly from the definition.(iv) Use dissections of the form

D = {brn, brn−1, brn−2, . . . , b}

with 0 < r and rn = a/b to compute∫ b

a

xm dx

directly from the definition.

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Q 19.16.⋆ If An and Gn are the arithmetic and geometric means of the npositive integers

n + 1, n + 2, . . . , n + n

show that, as n → ∞,

An

n→ 3

2and

Gn

n→ 4

e.

Deduce that e ≥ 8/3.

Q 19.17. ⋆ (i) Let

vn =

∫ (n+1)π

nπ

sin x

xdx.

By writing both integrals as integrals from 0 to π, or otherwise, show that|vn| ≥ |vn+1|. By using a theorem on the convergence of sums (to be stated)show that the sequence ∫ nπ

0

sin x

xdx → L

as n → ∞ through integer values of n where L is a strictly positive realnumber.

(ii) Deduce carefully that

∫ X

0

sin x

xdx → L

as X → ∞ through real values of X. Thus∫∞

0sin x

xdx exists with value L.

(iii)∗ Let

G(λ) =

∫ ∞

0

sin λx

xdx.

Show carefully (we have not actually proved a change of variables theoremfor infinite integrals) that G(λ) exists for all real λ and

G(λ) =

L if λ > 0

0 if λ = 0

−L if λ < 0

[Note that G is not continuous at 0. This is an indication of the unintuitivebehaviour which infinite integrals can exhibit.]

59

Q 19.18.⋆ (i) Suppose f1, f2 : [a, b] → R are increasing and g = f1 − f2.Show that there exists a K such that, whenever

a = x0 ≤ x1 ≤ x2 ≤ · · · ≤ xn = b

we haven∑

j=1

|g(xj) − g(xj−1)| ≤ K.

(ii) Let g : [−1, 1] → R be given by g(x) = x2 sin x−4 for x 6= 0, g(0) = 0.Show that g is once differentiable everywhere but that g is not the differenceof two increasing functions.

Final Note To Supervisors

Let me reiterate my request for corrections and improvements particularlyto the exercises. The easiest path for me is e-mail. My e-mail address istwk@dpmms.

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